math

An object that is falling or vertically projected into the air has it height, in feet, above the ground given by the formula below, where s is the height in feet, Vo is the original velocity of the object, in feet per second, t is the time the object is in motion in seconds, and So is the height, in feet, from which the object is dropped or projected. The figure shows that a ball is thrown straight up from a rooftop at an original velocity of 24 feet per second from a height of 96 feet. The ball misses the rooftop on its way down. How high above the ground will the ball be 2 seconds after thrown?
s=16t^2+VottSo. Please help, I'm stuck

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asked by pam
  1. fix the typo, and plug in the numbers

    s = 96 + 24t - 16t^2
    when t=2,
    s(2) = 96+48-64 = 80

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    posted by Steve
  2. f***

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    posted by Anonymous

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