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How do i find the final velocity with 2 height variables and gravity

  1. Sep 8, 2012 #1
    1.A tennis ball is dropped from 1.3 m above the
    ground. It rebounds to a height of 0.956 m.
    With what velocity does it hit the ground?
    The acceleration of gravity is 9.8 m/s^2
    (Let
    down be negative.)
    Answer in units of m/s



    2.With what velocity does it leave the ground?
    Answer in units of m/s

    3.If the tennis ball were in contact with the
    ground for 0.0132 s, find the acceleration
    given to the tennis ball by the ground.
    Answer in units of m/s^2
     
    Last edited: Sep 8, 2012
  2. jcsd
  3. Sep 8, 2012 #2

    CAF123

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    Gold Member

    You must show your attempt at the questions first. This can be any calculations, thoughts, misunderstandings etc..
     
  4. Sep 8, 2012 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What have you tried? You know, perhaps, that 'change in speed is equal to acceleration times time'? That comes directly from the definition of "acceleration". Here, the ball is "dropped" so the initial speed is 0 and so the final speed, the speed as it hits the ground is gt or -9.8t where t is the time the ball takes to hit the ground.

    You can find the time, t, by using "s= (1/2)at^2+ vt+ s0" where a is the acceleration, v is the initial speed and s0 is the initial height. That's a bit more compicated formula but you should have seen it if you are asked to do a problem like this. As before, a= -g= -9.8 and v= 0. The initial height is s0= 1.3 m above the ground so when the ball hits the ground, s= 0.

    So you want to solve (1/2)(-9.8)t^2+ 1.3= 0 for t and put that into v= -9.8t to find the ball's velocity when it first hits the ground.

    The information that "It rebounds to a height of 0.956 m" is irrelevant to that but would allow you to do exactly the same thing to determine the velocity the second time the ball hits the ground.
     
  5. Sep 8, 2012 #4
    ok lemme try

    it travels 1.3 m downwards right??

    so it becomes -V2 -U2=2(-9.8)(-1.3)

    as U=0 u can find The value of V


    Now Let velocity with which it leaves ground be u
    At height 0.956m, Velocity is 0 Acceleration is -9.8(due to gravity)

    using above equation u can get the velocity with which it leaves ground


    Now U can find the change in momentum by assuming mass of body = m

    mv1(after hitting ground{Final momentum})-mv2(before hitting ground{initial momentum}) = change in momentum

    Now force/[itex]\Delta[/itex]time gives us momentum

    so m(v1-v2)=ma/[itex]\Delta[/itex]time(0.0132s)(assuming acceleration =a)

    solve u should get acceleration
     
  6. Sep 8, 2012 #5

    CAF123

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    Gold Member

    It's probably worth mentioning that parts 1 and 2 of your question can be solved by energy methods as well.
     
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