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Ball dropped out a truck window, find bounce height and dist

  1. Dec 10, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball weighing 0.2 lbs is dropped out the window of a pickup truck moving horizontally at 45 mph. If the C.O.R. (e) between the ball and the ground is equal to 0.6, find the height of the ball's first bounce and the horizontal distance it will travel before bouncing again.

    I believe i have the height of the bounce, but am stuck on how far it will travel horizontally. I'll put below what i have so far.

    The attempt at a solution:

    mgh = 1/2mv2
    Vf = Square rt(2gh)
    square rt(2(32.2)(5ft)
    =17.9 ft/s

    V1 = 0.6(17.9ft/s)
    = 10.76 ft/s

    1/2mv2 = mgh
    h=vi2 / 2g
    =10.762 / 2(32.2)
    =1.798 ft
     
  2. jcsd
  3. Dec 11, 2015 #2

    haruspex

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    You can get the bounce height a bit more easily, it's just the original times the square of the coefficient.
    What is the horizontal velocity after the bounce? What else do you need to figure out to find the distance to the next bounce?
     
  4. Dec 11, 2015 #3
    Now you have found v1 right. From now on

    1/2*g*t^2= v1 ===> with that you find the t value ( the time the ball will go upwards)

    Then find 2t ( the time ball will bounce upwards and fall back down)

    2t* 45 mph = the distance your ball travel before its second bounce. Hope it was helpful and i didnt write something wrong by mistake
     
  5. Dec 11, 2015 #4
    Does anyone know if I found the first height correctly? Another student told me that you just take the orignal height times e to get that. That seems to simple to me but I could be wrong. So in this case it would be 0.6 x 5ft = 3ft
     
  6. Dec 11, 2015 #5

    jbriggs444

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    A quick trip to google indicates that the coefficient of restitution is a ratio of energy, not of velocity. Post #2 assumed the opposite.
    Scratch that. I read too quickly. It is a ratio of speed.

    If energy goes down by a factor of 0.6 then potential energy at the top of the bounce must go down by a factor of 0.6 and height at the top of the bounce will go down by a factor of... what?

    Scratch that. Re-read post #2.
     
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