Question 120868
The path of a falling object is given by the function s=-16t^2+v0t+s0 where v0 represents the initial velocity in ft/sec and s0 represents the initial height. The variable t is time in seconds, and the s is the height of the object in feet. 
a) If a rock is thrown upward with an initial velocity of 32 feet per second from the top of a 40-foot building, write the height equation using this information.
s(t)=-16t^2+32t+40
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b) How high is the rock after 0.5 seconds?
s(0.5) = -16(0.5)^2+32(0.5)+40 = 50 ft.
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c) After how many seconds will the rock reach maximum height?
Maximum occurs when t = -b/2a = -32/(2*-16) = 1 second
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d)What is the maximum height?
Maximum height occurs at s(1) = -16+32+40 = 56
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Cheers,
Stan H.