Calculus
posted by Kait .
A girls throws a tennis ball straight into the air with a velocity of 64 feet/sec. If acceleration due to gravity is 32 ft/sec2, how many seconds after it leaves the girl's hand will it take the ball to reach its highest point? Assume the position at time t = 0 is 0 feet.
Ok so I have d(t)=the integral of 64 so d(t)=64t+C. But d(0)=0 so C must be 0.
This is as far as I've gotten because I can't figure out how to solve for t if I don't have d(t) and I can't find what the highest position of the ball would be. Usually it's when v(t)=0 but all I have for v(t) is 64 ft/sec. Can this be determined?

Calculus 
Reiny
a = 32 ft/sec^2
v = 32t + c
when t = 0 , v = 64
64 = 0 + c , so c = 64, and
v = 32t + 64
d = 16t^2 + 64t + k, but when t=0 , d = 0
0 = 0 + 0 + k
k = 0
d = 16t^2 + 64t
The vertex of this parabola will tell you the maximum height, and the t when that max occurs.
I assume you know how to find that vertex. 
Calculus 
Kait
So it would take 2 seconds to reach the maximum height? Because v(t)=0 is the max height and setting 32t+64=0 would give t=2

Calculus 
bobpursley
yes, 2 seconds.
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