Physics: a ball is thrown vertically upward with an initial velocity of 80 feet per second. the distance (in feet) of the ball from the ground after t seconds if s=80t16tt^2
a) draw the illustration
b) for what time interval is the ball more than 96 feet above the ground?
c) what is the maximum height of the ball?
d) after how many seconds does the ball reach the maximum height?
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2 answers

another great help!
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answered by steph 
b)
80t  16t^2 > 96
16t^2  80t + 96 < 0
t^2  5t + 6 < 0
(t2)(t3) < 0
so the ball is higher than 96 for
2 < t < 3, that is, between 2 and 3 seconds
c) s = 16(t^2  5t + 25/4  25/4)
= 16(t5/2)^2 + 100
I will leave it up to you to interpret that result. 👍
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answered by Reiny
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