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Why do I get negative height?


Negative sign of accelerationCan you help me solve a difficult kinematics problem?A ball is dropped from the roof of a building. How fast is it moving after 4.9 seconds? (Absolute value)Height of buildingKinematic Problem, two balls thrown down from building at different times, when will they meet and what distance will they hit at?Calculating initial velocities given trajectory parabolaMotion of tennis ball bouncingConceptually, why is acceleration due to gravity always negative?Does the acceleration due of gravity taken positive or negative matters?How to solve this projectile motion physics problem in a book I'm studying?













5












$begingroup$


A baseball is thrown from the roof $50.1^circ$ above the horizontal. The initial velocity is $11.5 mathrmm/s$.



I'm trying to find how high the ball goes.
I choose positive direction upwards.



$$
H = frac12g v_0y^2
$$



But I get a negative answer since $g$ is negative. Is $g$ an absolute value when working with energy?










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  • 3




    $begingroup$
    How did you come up with that equation?
    $endgroup$
    – JMac
    Apr 2 at 11:22










  • $begingroup$
    @JMac I got it from: $K_1 + U_1 = K_2 + U_2$.
    $endgroup$
    – user644361
    Apr 2 at 11:25















5












$begingroup$


A baseball is thrown from the roof $50.1^circ$ above the horizontal. The initial velocity is $11.5 mathrmm/s$.



I'm trying to find how high the ball goes.
I choose positive direction upwards.



$$
H = frac12g v_0y^2
$$



But I get a negative answer since $g$ is negative. Is $g$ an absolute value when working with energy?










share|cite|improve this question









New contributor




user644361 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 3




    $begingroup$
    How did you come up with that equation?
    $endgroup$
    – JMac
    Apr 2 at 11:22










  • $begingroup$
    @JMac I got it from: $K_1 + U_1 = K_2 + U_2$.
    $endgroup$
    – user644361
    Apr 2 at 11:25













5












5








5


1



$begingroup$


A baseball is thrown from the roof $50.1^circ$ above the horizontal. The initial velocity is $11.5 mathrmm/s$.



I'm trying to find how high the ball goes.
I choose positive direction upwards.



$$
H = frac12g v_0y^2
$$



But I get a negative answer since $g$ is negative. Is $g$ an absolute value when working with energy?










share|cite|improve this question









New contributor




user644361 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




A baseball is thrown from the roof $50.1^circ$ above the horizontal. The initial velocity is $11.5 mathrmm/s$.



I'm trying to find how high the ball goes.
I choose positive direction upwards.



$$
H = frac12g v_0y^2
$$



But I get a negative answer since $g$ is negative. Is $g$ an absolute value when working with energy?







homework-and-exercises newtonian-mechanics energy-conservation coordinate-systems projectile






share|cite|improve this question









New contributor




user644361 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




user644361 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Apr 2 at 19:47









Qmechanic

107k121991238




107k121991238






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asked Apr 2 at 11:16









user644361user644361

3014




3014




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user644361 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





user644361 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user644361 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 3




    $begingroup$
    How did you come up with that equation?
    $endgroup$
    – JMac
    Apr 2 at 11:22










  • $begingroup$
    @JMac I got it from: $K_1 + U_1 = K_2 + U_2$.
    $endgroup$
    – user644361
    Apr 2 at 11:25












  • 3




    $begingroup$
    How did you come up with that equation?
    $endgroup$
    – JMac
    Apr 2 at 11:22










  • $begingroup$
    @JMac I got it from: $K_1 + U_1 = K_2 + U_2$.
    $endgroup$
    – user644361
    Apr 2 at 11:25







3




3




$begingroup$
How did you come up with that equation?
$endgroup$
– JMac
Apr 2 at 11:22




$begingroup$
How did you come up with that equation?
$endgroup$
– JMac
Apr 2 at 11:22












$begingroup$
@JMac I got it from: $K_1 + U_1 = K_2 + U_2$.
$endgroup$
– user644361
Apr 2 at 11:25




$begingroup$
@JMac I got it from: $K_1 + U_1 = K_2 + U_2$.
$endgroup$
– user644361
Apr 2 at 11:25










5 Answers
5






active

oldest

votes


















17












$begingroup$

$g$ is always positive. The negative sign you might usually see comes from defining down as negative, but the value of $g$ is always positive. This is why you don't ever see absolute value signs and why your equation is actually correct.



To add more detail, the value of $g$ is just the magnitude of acceleration due to gravity near Earth's surface. It is given by $$g=fracGMR^2$$
where $M$ and $R$ are the mass and radius of the earth respectively, and $G$ is a constant. All of these values are positive, so $g$ is also positive.






share|cite|improve this answer









$endgroup$








  • 9




    $begingroup$
    While I agree with this answer, I would like to point out that this is a result of convention (in particular the convention to define $g$ as above) and not a fundamental mathematical truth. If you were to define $g$ differently, you'd get a different answer. Of course this is always true, but it's important to point out here because there is a alternate definition of $g$ that makes enough sense to cause this question to be asked.
    $endgroup$
    – DreamConspiracy
    Apr 2 at 18:09






  • 2




    $begingroup$
    @DreamConspiracy I suppose this is true, if you no longer want to define $g$ to be the magnitude of the acceleration.
    $endgroup$
    – Aaron Stevens
    Apr 2 at 19:29










  • $begingroup$
    yes. The only reason I even mention this is because I have seen people do it this way (and in some cases it very well might be convenient), and that this is not necessarily wrong
    $endgroup$
    – DreamConspiracy
    Apr 2 at 19:32


















5












$begingroup$

$g$ should always be positive. It is the magnitude of the gravitational field strength. If you choose vertical up as positive $y$, then the gravitational field (a vector) will be $-ghatj$. The gravitational potential energy change within a small vertical range will be
$$Delta U_g=mgDelta y$$
where $g$ is reasonably constant within the $Delta y$ range. For example if a 2 kg object is moved near the surface of the earth from $y_a= 2$ m to $y_b= 3$ m,
$$Delta U_g=mg(y_b-y_a)=19.6~mathrmJ.$$
If it is moved from $y_b$ to $y_a$
$$Delta U_g=mg(y_a-y_b)=-19.6~mathrmJ.$$






share|cite|improve this answer









$endgroup$




















    4












    $begingroup$

    The first issue is that it's thrown at an angle.



    As the question seems to assumes no air friction, the part of the velocity we care about for the answer is its vertical component (11.5 x sin 50.1 = 8.82 m/s). The horizontal component says how far it will travel horizontally, which we aren't being asked about.



    So the question is the same as asking about a ball thrown vertically upward at 8.82 m/s.



    Next, we have to make a choice which direction we call "positive". That is a completely free choice.



    • If we call "up" the + direction , then the ball starts with positive velocity +8.82 (because its initial velocity is in the direction we defined as +), and has an acceleration in the negative direction, with acceleration -9.8 (because gravity acts in the downward direction we defined as -).

    • If we call "down" the + direction , then the ball starts with negative velocity -8.82 (because its initial velocity is in the direction we defined as -), and has an acceleration in the positive direction, with acceleration +9.8 (because gravity acts in the upward direction we defined as +).

    We can choose either of these, and the answer will be the same.



    If we define "up" as the positive direction, which is probably the usual way most people would do it, then the ball starts with a positive velocity, and experiences a constant acceleration (deceleration) in the negative direction. (See how we use positive and negative directions and +/-g?). The ball reaches its highest point when its velocity reaches zero, because after that it starts to move downward again in a negative direction. We want to know how far its travelled in that time.



    We can solve this using equations of motion, or energy - the answer will be the same. I'll show both methods.



    Using equations of motion



    There are different ways to write the equation when we know start velocity (v1 which is 8.82), end velocity (v2 which is zero), and acceleration (a which is -9.8), and want to know distance (s). One easy way is to work out the time (t) it takes:



    v2 = v1 + at



    => 0 = 8.82 + (-9.8) t



    => t = 8.82 / 9.8 = 0.9 seconds



    How far did it travel?



    s = (v1 + v2) t / 2



    s = (8.82 + 0) x 0.9 / 2



    s = 8.82 x 0.9 /2 = 3.969



    The ball travelled 3.969 metres upwards at its highest point.



    Using energy



    Initial K.E. is m.v12/2 = m.8.822/2 = 38.89 m



    Final K.E. = 0 (when the ball is at its highest point, all the K.E. has been transformed into P.E., and it has no vertical velocity.)



    The gain in P.E. by raising a ball of weight 'm' by height 'h' is mgh.



    But we don't have to consider directions of motion really. We only have to consider energy. It starts with some KE and ends with none. It starts with some PE and ends with a higher gravitational PE because its position is higher within a gravitational field (however it got there)



    => mgh = 38.89m



    => g.h = 38.89



    => 9.8.h = 38.89



    => h = 38.89 / 9.8 = 3.969



    Same answer - the ball rises by 3.969 metres at its highest point.






    share|cite|improve this answer











    $endgroup$




















      3












      $begingroup$

      I assume you got to this equation by setting the potential energy at the highest point $E_textpot=mgH$ equal to the vertical kinetic energy $E_textkin=fracmv_y^22$. In the formula for the potential energy, $g$ is to be taken positive, as otherwise this would yield a negative potential energy (which is possible, but not correct in this context).




      If you want to see that consistent, derive the formula for the potential energy yourself. In that case $F_textgrav=mg$ since you take $g$ to be negative, and we are only considering the $z$ component. Then, since for lifting something up we need to apply a force opposite the gravitational force, we have
      $$E_textpot = int_0^h -F_textgrav ds = int_0^h -mg,ds = left. -mgsright|_s=0^h = -mgh$$






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        I just don't get it how you can willy-nilly change the direction of something in physics when you've defined a positive direction.
        $endgroup$
        – user644361
        Apr 2 at 11:26










      • $begingroup$
        The formula doesn't have absolute value signs for g either, $|g|$
        $endgroup$
        – user644361
        Apr 2 at 11:29










      • $begingroup$
        @user644361 Which has higher potential energy, the ball at h=1m or the ball at h=10m?
        $endgroup$
        – PM 2Ring
        Apr 2 at 11:32






      • 5




        $begingroup$
        You don't change it willy-nilly. This is why you should know where formulas come from. If you look at the derivation of the $mgh$ formula, you see that $g$ is taken to be positive, as is usually convention.
        $endgroup$
        – noah
        Apr 2 at 11:44






      • 2




        $begingroup$
        The work done by a change in potential energy is $W = -Delta U$. Note the minus sign in that equation. You have another minus sign because weight acts in the negative $y$ direction, in your sign convention. You can't do mechanics by just picking equations at random from a textbook and hoping everything will work out right - you have to understand what you are doing.
        $endgroup$
        – alephzero
        Apr 2 at 11:58


















      2












      $begingroup$

      The signs work out correctly as long as you are careful.



      If an object accelerates at a constant acceleration $a$ over a distance $s$ then the work done on the object is force times distance. But we have $F=ma$, so the work done on the object is $mas$. So if $K_1$ and $K_2$ are the kinetic energy of the object at the start and at the end of the period, then



      $K_2-K_1=mas$



      In the case of ballistic motion in a vertical direction, $K_1=frac 1 2 mv^2$ where $v$ is the initial velocity of the object. If the object reaches a maximum height of $h$ then $K_2=0$ when $s=h$, so we have



      $-frac1 2 mv^2 = mah$



      Substituting $a=-g$ and we have



      $-frac1 2 mv^2 = -mgh \ Rightarrow h = frac v^2 2g $






      share|cite|improve this answer









      $endgroup$













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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        17












        $begingroup$

        $g$ is always positive. The negative sign you might usually see comes from defining down as negative, but the value of $g$ is always positive. This is why you don't ever see absolute value signs and why your equation is actually correct.



        To add more detail, the value of $g$ is just the magnitude of acceleration due to gravity near Earth's surface. It is given by $$g=fracGMR^2$$
        where $M$ and $R$ are the mass and radius of the earth respectively, and $G$ is a constant. All of these values are positive, so $g$ is also positive.






        share|cite|improve this answer









        $endgroup$








        • 9




          $begingroup$
          While I agree with this answer, I would like to point out that this is a result of convention (in particular the convention to define $g$ as above) and not a fundamental mathematical truth. If you were to define $g$ differently, you'd get a different answer. Of course this is always true, but it's important to point out here because there is a alternate definition of $g$ that makes enough sense to cause this question to be asked.
          $endgroup$
          – DreamConspiracy
          Apr 2 at 18:09






        • 2




          $begingroup$
          @DreamConspiracy I suppose this is true, if you no longer want to define $g$ to be the magnitude of the acceleration.
          $endgroup$
          – Aaron Stevens
          Apr 2 at 19:29










        • $begingroup$
          yes. The only reason I even mention this is because I have seen people do it this way (and in some cases it very well might be convenient), and that this is not necessarily wrong
          $endgroup$
          – DreamConspiracy
          Apr 2 at 19:32















        17












        $begingroup$

        $g$ is always positive. The negative sign you might usually see comes from defining down as negative, but the value of $g$ is always positive. This is why you don't ever see absolute value signs and why your equation is actually correct.



        To add more detail, the value of $g$ is just the magnitude of acceleration due to gravity near Earth's surface. It is given by $$g=fracGMR^2$$
        where $M$ and $R$ are the mass and radius of the earth respectively, and $G$ is a constant. All of these values are positive, so $g$ is also positive.






        share|cite|improve this answer









        $endgroup$








        • 9




          $begingroup$
          While I agree with this answer, I would like to point out that this is a result of convention (in particular the convention to define $g$ as above) and not a fundamental mathematical truth. If you were to define $g$ differently, you'd get a different answer. Of course this is always true, but it's important to point out here because there is a alternate definition of $g$ that makes enough sense to cause this question to be asked.
          $endgroup$
          – DreamConspiracy
          Apr 2 at 18:09






        • 2




          $begingroup$
          @DreamConspiracy I suppose this is true, if you no longer want to define $g$ to be the magnitude of the acceleration.
          $endgroup$
          – Aaron Stevens
          Apr 2 at 19:29










        • $begingroup$
          yes. The only reason I even mention this is because I have seen people do it this way (and in some cases it very well might be convenient), and that this is not necessarily wrong
          $endgroup$
          – DreamConspiracy
          Apr 2 at 19:32













        17












        17








        17





        $begingroup$

        $g$ is always positive. The negative sign you might usually see comes from defining down as negative, but the value of $g$ is always positive. This is why you don't ever see absolute value signs and why your equation is actually correct.



        To add more detail, the value of $g$ is just the magnitude of acceleration due to gravity near Earth's surface. It is given by $$g=fracGMR^2$$
        where $M$ and $R$ are the mass and radius of the earth respectively, and $G$ is a constant. All of these values are positive, so $g$ is also positive.






        share|cite|improve this answer









        $endgroup$



        $g$ is always positive. The negative sign you might usually see comes from defining down as negative, but the value of $g$ is always positive. This is why you don't ever see absolute value signs and why your equation is actually correct.



        To add more detail, the value of $g$ is just the magnitude of acceleration due to gravity near Earth's surface. It is given by $$g=fracGMR^2$$
        where $M$ and $R$ are the mass and radius of the earth respectively, and $G$ is a constant. All of these values are positive, so $g$ is also positive.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 2 at 11:51









        Aaron StevensAaron Stevens

        14.5k42453




        14.5k42453







        • 9




          $begingroup$
          While I agree with this answer, I would like to point out that this is a result of convention (in particular the convention to define $g$ as above) and not a fundamental mathematical truth. If you were to define $g$ differently, you'd get a different answer. Of course this is always true, but it's important to point out here because there is a alternate definition of $g$ that makes enough sense to cause this question to be asked.
          $endgroup$
          – DreamConspiracy
          Apr 2 at 18:09






        • 2




          $begingroup$
          @DreamConspiracy I suppose this is true, if you no longer want to define $g$ to be the magnitude of the acceleration.
          $endgroup$
          – Aaron Stevens
          Apr 2 at 19:29










        • $begingroup$
          yes. The only reason I even mention this is because I have seen people do it this way (and in some cases it very well might be convenient), and that this is not necessarily wrong
          $endgroup$
          – DreamConspiracy
          Apr 2 at 19:32












        • 9




          $begingroup$
          While I agree with this answer, I would like to point out that this is a result of convention (in particular the convention to define $g$ as above) and not a fundamental mathematical truth. If you were to define $g$ differently, you'd get a different answer. Of course this is always true, but it's important to point out here because there is a alternate definition of $g$ that makes enough sense to cause this question to be asked.
          $endgroup$
          – DreamConspiracy
          Apr 2 at 18:09






        • 2




          $begingroup$
          @DreamConspiracy I suppose this is true, if you no longer want to define $g$ to be the magnitude of the acceleration.
          $endgroup$
          – Aaron Stevens
          Apr 2 at 19:29










        • $begingroup$
          yes. The only reason I even mention this is because I have seen people do it this way (and in some cases it very well might be convenient), and that this is not necessarily wrong
          $endgroup$
          – DreamConspiracy
          Apr 2 at 19:32







        9




        9




        $begingroup$
        While I agree with this answer, I would like to point out that this is a result of convention (in particular the convention to define $g$ as above) and not a fundamental mathematical truth. If you were to define $g$ differently, you'd get a different answer. Of course this is always true, but it's important to point out here because there is a alternate definition of $g$ that makes enough sense to cause this question to be asked.
        $endgroup$
        – DreamConspiracy
        Apr 2 at 18:09




        $begingroup$
        While I agree with this answer, I would like to point out that this is a result of convention (in particular the convention to define $g$ as above) and not a fundamental mathematical truth. If you were to define $g$ differently, you'd get a different answer. Of course this is always true, but it's important to point out here because there is a alternate definition of $g$ that makes enough sense to cause this question to be asked.
        $endgroup$
        – DreamConspiracy
        Apr 2 at 18:09




        2




        2




        $begingroup$
        @DreamConspiracy I suppose this is true, if you no longer want to define $g$ to be the magnitude of the acceleration.
        $endgroup$
        – Aaron Stevens
        Apr 2 at 19:29




        $begingroup$
        @DreamConspiracy I suppose this is true, if you no longer want to define $g$ to be the magnitude of the acceleration.
        $endgroup$
        – Aaron Stevens
        Apr 2 at 19:29












        $begingroup$
        yes. The only reason I even mention this is because I have seen people do it this way (and in some cases it very well might be convenient), and that this is not necessarily wrong
        $endgroup$
        – DreamConspiracy
        Apr 2 at 19:32




        $begingroup$
        yes. The only reason I even mention this is because I have seen people do it this way (and in some cases it very well might be convenient), and that this is not necessarily wrong
        $endgroup$
        – DreamConspiracy
        Apr 2 at 19:32











        5












        $begingroup$

        $g$ should always be positive. It is the magnitude of the gravitational field strength. If you choose vertical up as positive $y$, then the gravitational field (a vector) will be $-ghatj$. The gravitational potential energy change within a small vertical range will be
        $$Delta U_g=mgDelta y$$
        where $g$ is reasonably constant within the $Delta y$ range. For example if a 2 kg object is moved near the surface of the earth from $y_a= 2$ m to $y_b= 3$ m,
        $$Delta U_g=mg(y_b-y_a)=19.6~mathrmJ.$$
        If it is moved from $y_b$ to $y_a$
        $$Delta U_g=mg(y_a-y_b)=-19.6~mathrmJ.$$






        share|cite|improve this answer









        $endgroup$

















          5












          $begingroup$

          $g$ should always be positive. It is the magnitude of the gravitational field strength. If you choose vertical up as positive $y$, then the gravitational field (a vector) will be $-ghatj$. The gravitational potential energy change within a small vertical range will be
          $$Delta U_g=mgDelta y$$
          where $g$ is reasonably constant within the $Delta y$ range. For example if a 2 kg object is moved near the surface of the earth from $y_a= 2$ m to $y_b= 3$ m,
          $$Delta U_g=mg(y_b-y_a)=19.6~mathrmJ.$$
          If it is moved from $y_b$ to $y_a$
          $$Delta U_g=mg(y_a-y_b)=-19.6~mathrmJ.$$






          share|cite|improve this answer









          $endgroup$















            5












            5








            5





            $begingroup$

            $g$ should always be positive. It is the magnitude of the gravitational field strength. If you choose vertical up as positive $y$, then the gravitational field (a vector) will be $-ghatj$. The gravitational potential energy change within a small vertical range will be
            $$Delta U_g=mgDelta y$$
            where $g$ is reasonably constant within the $Delta y$ range. For example if a 2 kg object is moved near the surface of the earth from $y_a= 2$ m to $y_b= 3$ m,
            $$Delta U_g=mg(y_b-y_a)=19.6~mathrmJ.$$
            If it is moved from $y_b$ to $y_a$
            $$Delta U_g=mg(y_a-y_b)=-19.6~mathrmJ.$$






            share|cite|improve this answer









            $endgroup$



            $g$ should always be positive. It is the magnitude of the gravitational field strength. If you choose vertical up as positive $y$, then the gravitational field (a vector) will be $-ghatj$. The gravitational potential energy change within a small vertical range will be
            $$Delta U_g=mgDelta y$$
            where $g$ is reasonably constant within the $Delta y$ range. For example if a 2 kg object is moved near the surface of the earth from $y_a= 2$ m to $y_b= 3$ m,
            $$Delta U_g=mg(y_b-y_a)=19.6~mathrmJ.$$
            If it is moved from $y_b$ to $y_a$
            $$Delta U_g=mg(y_a-y_b)=-19.6~mathrmJ.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Apr 2 at 12:04









            Bill NBill N

            9,87912241




            9,87912241





















                4












                $begingroup$

                The first issue is that it's thrown at an angle.



                As the question seems to assumes no air friction, the part of the velocity we care about for the answer is its vertical component (11.5 x sin 50.1 = 8.82 m/s). The horizontal component says how far it will travel horizontally, which we aren't being asked about.



                So the question is the same as asking about a ball thrown vertically upward at 8.82 m/s.



                Next, we have to make a choice which direction we call "positive". That is a completely free choice.



                • If we call "up" the + direction , then the ball starts with positive velocity +8.82 (because its initial velocity is in the direction we defined as +), and has an acceleration in the negative direction, with acceleration -9.8 (because gravity acts in the downward direction we defined as -).

                • If we call "down" the + direction , then the ball starts with negative velocity -8.82 (because its initial velocity is in the direction we defined as -), and has an acceleration in the positive direction, with acceleration +9.8 (because gravity acts in the upward direction we defined as +).

                We can choose either of these, and the answer will be the same.



                If we define "up" as the positive direction, which is probably the usual way most people would do it, then the ball starts with a positive velocity, and experiences a constant acceleration (deceleration) in the negative direction. (See how we use positive and negative directions and +/-g?). The ball reaches its highest point when its velocity reaches zero, because after that it starts to move downward again in a negative direction. We want to know how far its travelled in that time.



                We can solve this using equations of motion, or energy - the answer will be the same. I'll show both methods.



                Using equations of motion



                There are different ways to write the equation when we know start velocity (v1 which is 8.82), end velocity (v2 which is zero), and acceleration (a which is -9.8), and want to know distance (s). One easy way is to work out the time (t) it takes:



                v2 = v1 + at



                => 0 = 8.82 + (-9.8) t



                => t = 8.82 / 9.8 = 0.9 seconds



                How far did it travel?



                s = (v1 + v2) t / 2



                s = (8.82 + 0) x 0.9 / 2



                s = 8.82 x 0.9 /2 = 3.969



                The ball travelled 3.969 metres upwards at its highest point.



                Using energy



                Initial K.E. is m.v12/2 = m.8.822/2 = 38.89 m



                Final K.E. = 0 (when the ball is at its highest point, all the K.E. has been transformed into P.E., and it has no vertical velocity.)



                The gain in P.E. by raising a ball of weight 'm' by height 'h' is mgh.



                But we don't have to consider directions of motion really. We only have to consider energy. It starts with some KE and ends with none. It starts with some PE and ends with a higher gravitational PE because its position is higher within a gravitational field (however it got there)



                => mgh = 38.89m



                => g.h = 38.89



                => 9.8.h = 38.89



                => h = 38.89 / 9.8 = 3.969



                Same answer - the ball rises by 3.969 metres at its highest point.






                share|cite|improve this answer











                $endgroup$

















                  4












                  $begingroup$

                  The first issue is that it's thrown at an angle.



                  As the question seems to assumes no air friction, the part of the velocity we care about for the answer is its vertical component (11.5 x sin 50.1 = 8.82 m/s). The horizontal component says how far it will travel horizontally, which we aren't being asked about.



                  So the question is the same as asking about a ball thrown vertically upward at 8.82 m/s.



                  Next, we have to make a choice which direction we call "positive". That is a completely free choice.



                  • If we call "up" the + direction , then the ball starts with positive velocity +8.82 (because its initial velocity is in the direction we defined as +), and has an acceleration in the negative direction, with acceleration -9.8 (because gravity acts in the downward direction we defined as -).

                  • If we call "down" the + direction , then the ball starts with negative velocity -8.82 (because its initial velocity is in the direction we defined as -), and has an acceleration in the positive direction, with acceleration +9.8 (because gravity acts in the upward direction we defined as +).

                  We can choose either of these, and the answer will be the same.



                  If we define "up" as the positive direction, which is probably the usual way most people would do it, then the ball starts with a positive velocity, and experiences a constant acceleration (deceleration) in the negative direction. (See how we use positive and negative directions and +/-g?). The ball reaches its highest point when its velocity reaches zero, because after that it starts to move downward again in a negative direction. We want to know how far its travelled in that time.



                  We can solve this using equations of motion, or energy - the answer will be the same. I'll show both methods.



                  Using equations of motion



                  There are different ways to write the equation when we know start velocity (v1 which is 8.82), end velocity (v2 which is zero), and acceleration (a which is -9.8), and want to know distance (s). One easy way is to work out the time (t) it takes:



                  v2 = v1 + at



                  => 0 = 8.82 + (-9.8) t



                  => t = 8.82 / 9.8 = 0.9 seconds



                  How far did it travel?



                  s = (v1 + v2) t / 2



                  s = (8.82 + 0) x 0.9 / 2



                  s = 8.82 x 0.9 /2 = 3.969



                  The ball travelled 3.969 metres upwards at its highest point.



                  Using energy



                  Initial K.E. is m.v12/2 = m.8.822/2 = 38.89 m



                  Final K.E. = 0 (when the ball is at its highest point, all the K.E. has been transformed into P.E., and it has no vertical velocity.)



                  The gain in P.E. by raising a ball of weight 'm' by height 'h' is mgh.



                  But we don't have to consider directions of motion really. We only have to consider energy. It starts with some KE and ends with none. It starts with some PE and ends with a higher gravitational PE because its position is higher within a gravitational field (however it got there)



                  => mgh = 38.89m



                  => g.h = 38.89



                  => 9.8.h = 38.89



                  => h = 38.89 / 9.8 = 3.969



                  Same answer - the ball rises by 3.969 metres at its highest point.






                  share|cite|improve this answer











                  $endgroup$















                    4












                    4








                    4





                    $begingroup$

                    The first issue is that it's thrown at an angle.



                    As the question seems to assumes no air friction, the part of the velocity we care about for the answer is its vertical component (11.5 x sin 50.1 = 8.82 m/s). The horizontal component says how far it will travel horizontally, which we aren't being asked about.



                    So the question is the same as asking about a ball thrown vertically upward at 8.82 m/s.



                    Next, we have to make a choice which direction we call "positive". That is a completely free choice.



                    • If we call "up" the + direction , then the ball starts with positive velocity +8.82 (because its initial velocity is in the direction we defined as +), and has an acceleration in the negative direction, with acceleration -9.8 (because gravity acts in the downward direction we defined as -).

                    • If we call "down" the + direction , then the ball starts with negative velocity -8.82 (because its initial velocity is in the direction we defined as -), and has an acceleration in the positive direction, with acceleration +9.8 (because gravity acts in the upward direction we defined as +).

                    We can choose either of these, and the answer will be the same.



                    If we define "up" as the positive direction, which is probably the usual way most people would do it, then the ball starts with a positive velocity, and experiences a constant acceleration (deceleration) in the negative direction. (See how we use positive and negative directions and +/-g?). The ball reaches its highest point when its velocity reaches zero, because after that it starts to move downward again in a negative direction. We want to know how far its travelled in that time.



                    We can solve this using equations of motion, or energy - the answer will be the same. I'll show both methods.



                    Using equations of motion



                    There are different ways to write the equation when we know start velocity (v1 which is 8.82), end velocity (v2 which is zero), and acceleration (a which is -9.8), and want to know distance (s). One easy way is to work out the time (t) it takes:



                    v2 = v1 + at



                    => 0 = 8.82 + (-9.8) t



                    => t = 8.82 / 9.8 = 0.9 seconds



                    How far did it travel?



                    s = (v1 + v2) t / 2



                    s = (8.82 + 0) x 0.9 / 2



                    s = 8.82 x 0.9 /2 = 3.969



                    The ball travelled 3.969 metres upwards at its highest point.



                    Using energy



                    Initial K.E. is m.v12/2 = m.8.822/2 = 38.89 m



                    Final K.E. = 0 (when the ball is at its highest point, all the K.E. has been transformed into P.E., and it has no vertical velocity.)



                    The gain in P.E. by raising a ball of weight 'm' by height 'h' is mgh.



                    But we don't have to consider directions of motion really. We only have to consider energy. It starts with some KE and ends with none. It starts with some PE and ends with a higher gravitational PE because its position is higher within a gravitational field (however it got there)



                    => mgh = 38.89m



                    => g.h = 38.89



                    => 9.8.h = 38.89



                    => h = 38.89 / 9.8 = 3.969



                    Same answer - the ball rises by 3.969 metres at its highest point.






                    share|cite|improve this answer











                    $endgroup$



                    The first issue is that it's thrown at an angle.



                    As the question seems to assumes no air friction, the part of the velocity we care about for the answer is its vertical component (11.5 x sin 50.1 = 8.82 m/s). The horizontal component says how far it will travel horizontally, which we aren't being asked about.



                    So the question is the same as asking about a ball thrown vertically upward at 8.82 m/s.



                    Next, we have to make a choice which direction we call "positive". That is a completely free choice.



                    • If we call "up" the + direction , then the ball starts with positive velocity +8.82 (because its initial velocity is in the direction we defined as +), and has an acceleration in the negative direction, with acceleration -9.8 (because gravity acts in the downward direction we defined as -).

                    • If we call "down" the + direction , then the ball starts with negative velocity -8.82 (because its initial velocity is in the direction we defined as -), and has an acceleration in the positive direction, with acceleration +9.8 (because gravity acts in the upward direction we defined as +).

                    We can choose either of these, and the answer will be the same.



                    If we define "up" as the positive direction, which is probably the usual way most people would do it, then the ball starts with a positive velocity, and experiences a constant acceleration (deceleration) in the negative direction. (See how we use positive and negative directions and +/-g?). The ball reaches its highest point when its velocity reaches zero, because after that it starts to move downward again in a negative direction. We want to know how far its travelled in that time.



                    We can solve this using equations of motion, or energy - the answer will be the same. I'll show both methods.



                    Using equations of motion



                    There are different ways to write the equation when we know start velocity (v1 which is 8.82), end velocity (v2 which is zero), and acceleration (a which is -9.8), and want to know distance (s). One easy way is to work out the time (t) it takes:



                    v2 = v1 + at



                    => 0 = 8.82 + (-9.8) t



                    => t = 8.82 / 9.8 = 0.9 seconds



                    How far did it travel?



                    s = (v1 + v2) t / 2



                    s = (8.82 + 0) x 0.9 / 2



                    s = 8.82 x 0.9 /2 = 3.969



                    The ball travelled 3.969 metres upwards at its highest point.



                    Using energy



                    Initial K.E. is m.v12/2 = m.8.822/2 = 38.89 m



                    Final K.E. = 0 (when the ball is at its highest point, all the K.E. has been transformed into P.E., and it has no vertical velocity.)



                    The gain in P.E. by raising a ball of weight 'm' by height 'h' is mgh.



                    But we don't have to consider directions of motion really. We only have to consider energy. It starts with some KE and ends with none. It starts with some PE and ends with a higher gravitational PE because its position is higher within a gravitational field (however it got there)



                    => mgh = 38.89m



                    => g.h = 38.89



                    => 9.8.h = 38.89



                    => h = 38.89 / 9.8 = 3.969



                    Same answer - the ball rises by 3.969 metres at its highest point.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Apr 2 at 20:45

























                    answered Apr 2 at 19:41









                    StilezStilez

                    1,468413




                    1,468413





















                        3












                        $begingroup$

                        I assume you got to this equation by setting the potential energy at the highest point $E_textpot=mgH$ equal to the vertical kinetic energy $E_textkin=fracmv_y^22$. In the formula for the potential energy, $g$ is to be taken positive, as otherwise this would yield a negative potential energy (which is possible, but not correct in this context).




                        If you want to see that consistent, derive the formula for the potential energy yourself. In that case $F_textgrav=mg$ since you take $g$ to be negative, and we are only considering the $z$ component. Then, since for lifting something up we need to apply a force opposite the gravitational force, we have
                        $$E_textpot = int_0^h -F_textgrav ds = int_0^h -mg,ds = left. -mgsright|_s=0^h = -mgh$$






                        share|cite|improve this answer











                        $endgroup$












                        • $begingroup$
                          I just don't get it how you can willy-nilly change the direction of something in physics when you've defined a positive direction.
                          $endgroup$
                          – user644361
                          Apr 2 at 11:26










                        • $begingroup$
                          The formula doesn't have absolute value signs for g either, $|g|$
                          $endgroup$
                          – user644361
                          Apr 2 at 11:29










                        • $begingroup$
                          @user644361 Which has higher potential energy, the ball at h=1m or the ball at h=10m?
                          $endgroup$
                          – PM 2Ring
                          Apr 2 at 11:32






                        • 5




                          $begingroup$
                          You don't change it willy-nilly. This is why you should know where formulas come from. If you look at the derivation of the $mgh$ formula, you see that $g$ is taken to be positive, as is usually convention.
                          $endgroup$
                          – noah
                          Apr 2 at 11:44






                        • 2




                          $begingroup$
                          The work done by a change in potential energy is $W = -Delta U$. Note the minus sign in that equation. You have another minus sign because weight acts in the negative $y$ direction, in your sign convention. You can't do mechanics by just picking equations at random from a textbook and hoping everything will work out right - you have to understand what you are doing.
                          $endgroup$
                          – alephzero
                          Apr 2 at 11:58















                        3












                        $begingroup$

                        I assume you got to this equation by setting the potential energy at the highest point $E_textpot=mgH$ equal to the vertical kinetic energy $E_textkin=fracmv_y^22$. In the formula for the potential energy, $g$ is to be taken positive, as otherwise this would yield a negative potential energy (which is possible, but not correct in this context).




                        If you want to see that consistent, derive the formula for the potential energy yourself. In that case $F_textgrav=mg$ since you take $g$ to be negative, and we are only considering the $z$ component. Then, since for lifting something up we need to apply a force opposite the gravitational force, we have
                        $$E_textpot = int_0^h -F_textgrav ds = int_0^h -mg,ds = left. -mgsright|_s=0^h = -mgh$$






                        share|cite|improve this answer











                        $endgroup$












                        • $begingroup$
                          I just don't get it how you can willy-nilly change the direction of something in physics when you've defined a positive direction.
                          $endgroup$
                          – user644361
                          Apr 2 at 11:26










                        • $begingroup$
                          The formula doesn't have absolute value signs for g either, $|g|$
                          $endgroup$
                          – user644361
                          Apr 2 at 11:29










                        • $begingroup$
                          @user644361 Which has higher potential energy, the ball at h=1m or the ball at h=10m?
                          $endgroup$
                          – PM 2Ring
                          Apr 2 at 11:32






                        • 5




                          $begingroup$
                          You don't change it willy-nilly. This is why you should know where formulas come from. If you look at the derivation of the $mgh$ formula, you see that $g$ is taken to be positive, as is usually convention.
                          $endgroup$
                          – noah
                          Apr 2 at 11:44






                        • 2




                          $begingroup$
                          The work done by a change in potential energy is $W = -Delta U$. Note the minus sign in that equation. You have another minus sign because weight acts in the negative $y$ direction, in your sign convention. You can't do mechanics by just picking equations at random from a textbook and hoping everything will work out right - you have to understand what you are doing.
                          $endgroup$
                          – alephzero
                          Apr 2 at 11:58













                        3












                        3








                        3





                        $begingroup$

                        I assume you got to this equation by setting the potential energy at the highest point $E_textpot=mgH$ equal to the vertical kinetic energy $E_textkin=fracmv_y^22$. In the formula for the potential energy, $g$ is to be taken positive, as otherwise this would yield a negative potential energy (which is possible, but not correct in this context).




                        If you want to see that consistent, derive the formula for the potential energy yourself. In that case $F_textgrav=mg$ since you take $g$ to be negative, and we are only considering the $z$ component. Then, since for lifting something up we need to apply a force opposite the gravitational force, we have
                        $$E_textpot = int_0^h -F_textgrav ds = int_0^h -mg,ds = left. -mgsright|_s=0^h = -mgh$$






                        share|cite|improve this answer











                        $endgroup$



                        I assume you got to this equation by setting the potential energy at the highest point $E_textpot=mgH$ equal to the vertical kinetic energy $E_textkin=fracmv_y^22$. In the formula for the potential energy, $g$ is to be taken positive, as otherwise this would yield a negative potential energy (which is possible, but not correct in this context).




                        If you want to see that consistent, derive the formula for the potential energy yourself. In that case $F_textgrav=mg$ since you take $g$ to be negative, and we are only considering the $z$ component. Then, since for lifting something up we need to apply a force opposite the gravitational force, we have
                        $$E_textpot = int_0^h -F_textgrav ds = int_0^h -mg,ds = left. -mgsright|_s=0^h = -mgh$$







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Apr 2 at 11:58

























                        answered Apr 2 at 11:22









                        noahnoah

                        4,13311226




                        4,13311226











                        • $begingroup$
                          I just don't get it how you can willy-nilly change the direction of something in physics when you've defined a positive direction.
                          $endgroup$
                          – user644361
                          Apr 2 at 11:26










                        • $begingroup$
                          The formula doesn't have absolute value signs for g either, $|g|$
                          $endgroup$
                          – user644361
                          Apr 2 at 11:29










                        • $begingroup$
                          @user644361 Which has higher potential energy, the ball at h=1m or the ball at h=10m?
                          $endgroup$
                          – PM 2Ring
                          Apr 2 at 11:32






                        • 5




                          $begingroup$
                          You don't change it willy-nilly. This is why you should know where formulas come from. If you look at the derivation of the $mgh$ formula, you see that $g$ is taken to be positive, as is usually convention.
                          $endgroup$
                          – noah
                          Apr 2 at 11:44






                        • 2




                          $begingroup$
                          The work done by a change in potential energy is $W = -Delta U$. Note the minus sign in that equation. You have another minus sign because weight acts in the negative $y$ direction, in your sign convention. You can't do mechanics by just picking equations at random from a textbook and hoping everything will work out right - you have to understand what you are doing.
                          $endgroup$
                          – alephzero
                          Apr 2 at 11:58
















                        • $begingroup$
                          I just don't get it how you can willy-nilly change the direction of something in physics when you've defined a positive direction.
                          $endgroup$
                          – user644361
                          Apr 2 at 11:26










                        • $begingroup$
                          The formula doesn't have absolute value signs for g either, $|g|$
                          $endgroup$
                          – user644361
                          Apr 2 at 11:29










                        • $begingroup$
                          @user644361 Which has higher potential energy, the ball at h=1m or the ball at h=10m?
                          $endgroup$
                          – PM 2Ring
                          Apr 2 at 11:32






                        • 5




                          $begingroup$
                          You don't change it willy-nilly. This is why you should know where formulas come from. If you look at the derivation of the $mgh$ formula, you see that $g$ is taken to be positive, as is usually convention.
                          $endgroup$
                          – noah
                          Apr 2 at 11:44






                        • 2




                          $begingroup$
                          The work done by a change in potential energy is $W = -Delta U$. Note the minus sign in that equation. You have another minus sign because weight acts in the negative $y$ direction, in your sign convention. You can't do mechanics by just picking equations at random from a textbook and hoping everything will work out right - you have to understand what you are doing.
                          $endgroup$
                          – alephzero
                          Apr 2 at 11:58















                        $begingroup$
                        I just don't get it how you can willy-nilly change the direction of something in physics when you've defined a positive direction.
                        $endgroup$
                        – user644361
                        Apr 2 at 11:26




                        $begingroup$
                        I just don't get it how you can willy-nilly change the direction of something in physics when you've defined a positive direction.
                        $endgroup$
                        – user644361
                        Apr 2 at 11:26












                        $begingroup$
                        The formula doesn't have absolute value signs for g either, $|g|$
                        $endgroup$
                        – user644361
                        Apr 2 at 11:29




                        $begingroup$
                        The formula doesn't have absolute value signs for g either, $|g|$
                        $endgroup$
                        – user644361
                        Apr 2 at 11:29












                        $begingroup$
                        @user644361 Which has higher potential energy, the ball at h=1m or the ball at h=10m?
                        $endgroup$
                        – PM 2Ring
                        Apr 2 at 11:32




                        $begingroup$
                        @user644361 Which has higher potential energy, the ball at h=1m or the ball at h=10m?
                        $endgroup$
                        – PM 2Ring
                        Apr 2 at 11:32




                        5




                        5




                        $begingroup$
                        You don't change it willy-nilly. This is why you should know where formulas come from. If you look at the derivation of the $mgh$ formula, you see that $g$ is taken to be positive, as is usually convention.
                        $endgroup$
                        – noah
                        Apr 2 at 11:44




                        $begingroup$
                        You don't change it willy-nilly. This is why you should know where formulas come from. If you look at the derivation of the $mgh$ formula, you see that $g$ is taken to be positive, as is usually convention.
                        $endgroup$
                        – noah
                        Apr 2 at 11:44




                        2




                        2




                        $begingroup$
                        The work done by a change in potential energy is $W = -Delta U$. Note the minus sign in that equation. You have another minus sign because weight acts in the negative $y$ direction, in your sign convention. You can't do mechanics by just picking equations at random from a textbook and hoping everything will work out right - you have to understand what you are doing.
                        $endgroup$
                        – alephzero
                        Apr 2 at 11:58




                        $begingroup$
                        The work done by a change in potential energy is $W = -Delta U$. Note the minus sign in that equation. You have another minus sign because weight acts in the negative $y$ direction, in your sign convention. You can't do mechanics by just picking equations at random from a textbook and hoping everything will work out right - you have to understand what you are doing.
                        $endgroup$
                        – alephzero
                        Apr 2 at 11:58











                        2












                        $begingroup$

                        The signs work out correctly as long as you are careful.



                        If an object accelerates at a constant acceleration $a$ over a distance $s$ then the work done on the object is force times distance. But we have $F=ma$, so the work done on the object is $mas$. So if $K_1$ and $K_2$ are the kinetic energy of the object at the start and at the end of the period, then



                        $K_2-K_1=mas$



                        In the case of ballistic motion in a vertical direction, $K_1=frac 1 2 mv^2$ where $v$ is the initial velocity of the object. If the object reaches a maximum height of $h$ then $K_2=0$ when $s=h$, so we have



                        $-frac1 2 mv^2 = mah$



                        Substituting $a=-g$ and we have



                        $-frac1 2 mv^2 = -mgh \ Rightarrow h = frac v^2 2g $






                        share|cite|improve this answer









                        $endgroup$

















                          2












                          $begingroup$

                          The signs work out correctly as long as you are careful.



                          If an object accelerates at a constant acceleration $a$ over a distance $s$ then the work done on the object is force times distance. But we have $F=ma$, so the work done on the object is $mas$. So if $K_1$ and $K_2$ are the kinetic energy of the object at the start and at the end of the period, then



                          $K_2-K_1=mas$



                          In the case of ballistic motion in a vertical direction, $K_1=frac 1 2 mv^2$ where $v$ is the initial velocity of the object. If the object reaches a maximum height of $h$ then $K_2=0$ when $s=h$, so we have



                          $-frac1 2 mv^2 = mah$



                          Substituting $a=-g$ and we have



                          $-frac1 2 mv^2 = -mgh \ Rightarrow h = frac v^2 2g $






                          share|cite|improve this answer









                          $endgroup$















                            2












                            2








                            2





                            $begingroup$

                            The signs work out correctly as long as you are careful.



                            If an object accelerates at a constant acceleration $a$ over a distance $s$ then the work done on the object is force times distance. But we have $F=ma$, so the work done on the object is $mas$. So if $K_1$ and $K_2$ are the kinetic energy of the object at the start and at the end of the period, then



                            $K_2-K_1=mas$



                            In the case of ballistic motion in a vertical direction, $K_1=frac 1 2 mv^2$ where $v$ is the initial velocity of the object. If the object reaches a maximum height of $h$ then $K_2=0$ when $s=h$, so we have



                            $-frac1 2 mv^2 = mah$



                            Substituting $a=-g$ and we have



                            $-frac1 2 mv^2 = -mgh \ Rightarrow h = frac v^2 2g $






                            share|cite|improve this answer









                            $endgroup$



                            The signs work out correctly as long as you are careful.



                            If an object accelerates at a constant acceleration $a$ over a distance $s$ then the work done on the object is force times distance. But we have $F=ma$, so the work done on the object is $mas$. So if $K_1$ and $K_2$ are the kinetic energy of the object at the start and at the end of the period, then



                            $K_2-K_1=mas$



                            In the case of ballistic motion in a vertical direction, $K_1=frac 1 2 mv^2$ where $v$ is the initial velocity of the object. If the object reaches a maximum height of $h$ then $K_2=0$ when $s=h$, so we have



                            $-frac1 2 mv^2 = mah$



                            Substituting $a=-g$ and we have



                            $-frac1 2 mv^2 = -mgh \ Rightarrow h = frac v^2 2g $







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Apr 2 at 12:13









                            gandalf61gandalf61

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