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put this solution on YOUR website!A ball is thrown vertically upward from the ground with an initial speed of 80 ft/sec. Its height is given by h=80t-16t^2. How high does the ball go? When does the ball hit the ground?
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h(t) = 80t-16t^2
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maximum height occurs when t = -b/2a = -80/(2*-16) = 5/2 = 2.5 seconds
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The height when t=2.5 sec is t(2.5) = 80*2.5 - 16*(2.5)^2 = 100 ft.
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The ball hits the ground when the h(t) = 0
80t - 16t^2 = 0
16t(5-t) = 0
t= 0 (it starts on the ground)
or
t= 5 seconds (it is back on the ground in 5 seconds)
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Cheers,
Stan H.
