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Kinematics- ball thrown

  1. Sep 19, 2006 #1
    kinematics-- ball thrown

    A ball is thrown upward from the edge of a building that is 100m tall. On its way back down, the ball just misses the building, falling to the ground below. The elapsed time between the throw and the ball hitting the ground is 6.5 sec.

    A) The initial speed of the ball
    B) Position and speed of the ball after 1 sec and after 4 secs.
    C) The maximum height reached by the ball, measured above the ground.

    *Ok, I've worked a few problems where the object is dropped directly from the top of the building but none where the object is actually thrown upward!*
  2. jcsd
  3. Sep 19, 2006 #2
    Well u just hav 2 use the same equations of motion

    The usual convention for these 1s is up is positive and down is negative so the initial speed u is a positive num since in the up direction. The displacement s will be -100 and g is -9.81. Putting in the numbers:

    s = u*t - 0.5*g*t^2
    -100 = 6.5*u - 0.5*(9.81)*(6.5)^2
    -100 = 6.5*u - 207.236
    6.5*u = 107.236
    u = 16.498

    To get the speeds just use v = u - g*t
    positions: s = u*t - 0.5*g*t^2

    U can use the fact that v becomes 0 at the max height to find it:
    v^2 = u^2 - 2*g*s
    u^2 = 2*g*s
    s = u^2/(2*g)
    Last edited: Sep 19, 2006
  4. Sep 19, 2006 #3

    why is s -100? I thought it should be +100 ?

    d=v0*t + 0.5at^2 isn't that the correct formula?
  5. Sep 20, 2006 #4

    Andrew Mason

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    Yes. But a = is negative if up is positive. Final d (vertical displacement) has to have the opposite sign of v0 and v0 has to have the opposite sign of a since v and a are in opposite directions.

    I would suggest that you plot a graph of velocity vs. time and the up direction positive. The graph is a straight line with slope -9.8 m/sec^2.

    [Of course, you could also do it with the down direction being positive and the slope + g but then the initial velocity would be negative.]


    PS and do not double post!!
    Last edited: Sep 20, 2006
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