# Calculus

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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 -

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 -

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 -

yes, 2 seconds.

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