ball is thrown up at 88 ft/s from a height of 25 feet. Path of ball is
y = -16t^2 + 88t +25. y is the height of ball above the ground t seconds after being thrown. when does it reach maxiumum height? (I got that at 2.75 s)

At what time does the ball hit the ground? (I set equation to =0, but I got 5.2 s and answer is 5.77 s)

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  2. 1
asked by BJ
  1. Vf = Vo(t) - 32(t)
    Vf = 88 - 32t or t = 2.75sec.

    Height reached from 25 ft. height
    h = 88(t) - 32(t^2)/2
    h = 88(2.75) - 16(2.75)^2 = 121 ft.

    It takes the same 2.75 sec. to return to the original launch height of 25 feet and an increment more to fall the 25 feet further to the ground.

    h = Vo(t) + 32(t^2)/2
    (121 + 25) = 0(t) + 16(t^2)
    t - sqrt(146/16) = 3.02

    The time from launch to impact is therefore (2.75 + 3.02) = 5.77 sec.

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    posted by tchrwill

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