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What was the total time of fall?

  1. Aug 25, 2008 #1
    A ball, dropped from rest, covers three-sevenths of the distance to the ground in the last two seconds of its fall.

    (a) From what height was the ball dropped?
    _m
    (b) What was the total time of fall?
    _s

    For a, don't i need the final velocity to solve the problem? Vi=o A=-9.81?
    2 days ago - 1 day left to answer.
     
  2. jcsd
  3. Aug 25, 2008 #2

    HallsofIvy

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    Re: Kinematics!

    No, you don't need the final velocity (although you can calculate it after you have solved b). I presume you know that the height of the ball after t seconds is s= h- (1/2)At2= h- 4.905t2. It will hit the ground when s= h- 4.905t2= 0 so the time it hits the ground, T, is [itex]T= \sqrt{h/4.905}[/itex].
    the ball "covers three-sevenths of the distance to the ground in the last two seconds of its fall." which tells us that [itex]s(T- 2)= h- 4.905(\sqrt{h/4.905}- 2)^2= (3/7) h[/itex].
    Solve that equation for h.
     
  4. Aug 25, 2008 #3
    Re: Kinematics!



    [itex]s(T- 2)= h- 4.905(\sqrt{h/4.905}- 2)^2= (3/7) h[/itex]. For that part, i just do [itex]h- 4.905(\sqrt{h/4.905}- 2)^2= (3/7) h[/itex] that and solve for h?

    i got 45.78 for h, but it didn't work out ;(
     
    Last edited: Aug 25, 2008
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