1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Ball Is Thrown Vertically Upwards, Find The Ratio of PE to KE

  1. Aug 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball of mass 65 g is thrown vertically upwards from ground level with a speed of 16 m s–1. Air resistance is negligible. The ball takes time t to reach maximum height. For time t 2 after the ball has been thrown, calculate the ratio of the potential energy of ball to the kinetic energy of ball .


    2. Relevant equations
    T=(usinA)/g, where A=90
    s= 1/2at^2 + ut
    GPE=mgh
    KE=1/2mv^2
    3. The attempt at a solution
    T=(usinA)/g
    = 16/9.81
    = 1.63s
    so t/2= 0.816
    at 0.816s,
    s= (1/2 x 9.81 x 0.816^2) + (16 x 0.816)
    = 16.322m
    GPE=mgh
    0.065 x 9.81 x 16.322
    KE= 1/2mv^2
    1/2 x 0.065 x v^2
    v^2-u^2=2as
    v^2=2 x 9.81 x 16.322
    substituting,
    KE= 1/2 x 0.065 x 2 x 9.81 x 16.322
    = 0.065 x 9.81 x 16.322
    So, GPE:KE
    = (0.065 x 9.81 x 16.322) / (0.065 x 9.81 x 16.322)
    =1:1
    The answer is incorrect. The correct answer is 3:1.
     
    Last edited: Aug 18, 2015
  2. jcsd
  3. Aug 18, 2015 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Be careful with signs. Are you defining positive as up or down? Either way, make sure you are consistent on that with all accelerations, velocities and distances.
     
  4. Aug 18, 2015 #3

    andrevdh

    User Avatar
    Homework Helper

    Try to work in symbols right up to the very end.
     

    Attached Files:

  5. Aug 18, 2015 #4
    All the calculations I did dealt with the ball moving upwards till it reached it's maximum height. So I took the upwards direction to be positive.
     
  6. Aug 18, 2015 #5
    But how would I find the ratio between the two? Shouldn't I have divided the two?
    Yeah, working in symbols would have made it easier.
     
  7. Aug 18, 2015 #6

    andrevdh

    User Avatar
    Homework Helper

    So why don't you try that if it is easier?
     

    Attached Files:

  8. Aug 18, 2015 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    So what sign should be on the 9.81 gravitational acceleration?
     
  9. Aug 19, 2015 #8

    andrevdh

    User Avatar
    Homework Helper

    Your error is at this point as haruspex mentions.
    I seems to get a ratio of 4:1?
     
    Last edited: Aug 19, 2015
  10. Aug 20, 2015 #9
    It should be negative? But what should I do after getting s= -16.322m?
     
  11. Aug 20, 2015 #10
    4:1? The answer says it is 3:1.
    In their working they have got (3/4) / (1/4). So the 4s cancel. I didn't get any of these numbers.
     
  12. Aug 20, 2015 #11
    Here's what I would do. Calculate the speed of the ball at the time given. Use that to find the KE of the ball. Then, find the height. Convert that to PE and find the ratio.
     
  13. Aug 20, 2015 #12

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I didn't say change both signs. It's only the 9.81 that had the wrong sign. The distance must be positive.
     
  14. Aug 20, 2015 #13

    jbriggs444

    User Avatar
    Science Advisor

    You should not be negating the sign on s. You should be negating your sign on a and then recalculating s. What do you get for s now?
    Edit: didn't see haruspex in there. Please pardon the duplication.
     
  15. Aug 20, 2015 #14

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Note that the answer is independent of ##g##, ##m## and ##u## and, in fact, there was no need for any numerical calculations.
     
  16. Aug 20, 2015 #15
    is this a trick question?

    The ball takes time t to reach maximum height. For time t 2... (twice the time it takes to reach max height?)

    ie. the ball is back on the ground so its PE and KE are both zero....unless its rolling around...
     
  17. Aug 20, 2015 #16

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    If you study the working in the OP you can deduce it was meant to say t/2.
     
  18. Aug 20, 2015 #17
    ah! I only read the question..the working looked like too much of a mess

    yes, it is 3:1
     
  19. Aug 21, 2015 #18
    Okay, then--
    s= (1/2 x -9.81 x 0.816^2) + (16 x 0.816)
    = 9.789
    GPE=mgh
    Then, 0.065 x -9.81 x 9.789
    KE= 1/2mv^2
    v^2-u^2=2as
    v^2=2 x -9.81 x 9.789
    KE= 1/2 x 2 x -9.81 x 9.789 x 0.065
    GPE:KE
    (0.065 x -9.81 x 9.789)/ (1/2 x 2 x -9.81 x 9.789 x 0.065)
    = 1:1
    The answer still doesn't seem to match.
     
  20. Aug 21, 2015 #19
    I'm so sorry, it is supposed to be t/2!
     
  21. Aug 21, 2015 #20
    you can do it the long way:

    total energy at time of ball release is KE

    KE = 1/2 mv^2 = 8.32 J

    Time to reach maximum height =
    t = u / -g = 1.63s

    Half this time is 0.815s

    v = u + a * 0.815 = 8 (note this, it is half initial speed)


    KE at this t - 0.815

    1/2 mv^2 = 2.08 J

    which is a quarter of the intial KE (note this, it is quarter initial KE)


    The total energy is 8.32J
    at t/2
    KE = 2.08J and PE is the remainder 6.24J


    so the ratio is 3:1

    The bits on bold allude to the fact that you did not have to calulate anything.
     
    Last edited: Aug 21, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: A Ball Is Thrown Vertically Upwards, Find The Ratio of PE to KE
Loading...