calculus
posted by Ashley .
A ball is dropped from the top of a building. The height, , of the ball above the ground (in feet) is given as a function of time, , (in seconds) by
y = 1640  16t^2
y'= 32t
When does the ball hit the ground, and how fast is it going at that time? Give your answer in feet per second and in miles per hour (1 ft/sec = 15/22 mph).
v(t) = mph
v(t) = ft per second

for the first part solve
0 = 1640  16t^2
16t^2 = 1640
t^2 = 102.5
t = 10.124 seconds
sub that into y' to get the velocity in ft/sec
Use the formula given to you to change to mph
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