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Pre-Calculus

A ball is thrown vertically upward with an initial velocity of 64 feet per second. The distance(in feet) of the ball from the ground after t seconds is s=64t-16t.

At what time will the ball strike the ground?

For what time t is the ball more than 28 feet above the ground.

The ball will strike the ground when t is _____ seconds.

The ball is more than 28 feet above the ground for the time t when ____<t<____.

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Cicley

  1. you should proably first simplify 64t-16t = 48t. Then, 48t=s. Plug it in, see what you get.

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    mystify

  2. s = 64t - 16t^2
    s=0 at t= 0,2

    ball is more than 28 ft high when

    64t - 16t^2 > 28
    -16t^2 + 64t - 28 > 0
    -4(2t-1)(2t-7) > 0

    1/2 < t < 7/2

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    Steve

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