Answer:
64 feet
Explanation:
Complete question
A ball is thrown straight up with initial velocity of 64 ft./s. The height of the ball t seconds after it is thrown is given by the formula s(t)=64t-16t^2. Find the maximum height reached by the ball.
At maximum height, the velocity of the ball is zero, i.e ds/dt =0
ds/dt = 64 - 32t
64 - 32t = 0
32t = 64
t = 64/32
t = 2secs
Substitute t = 2secs into the given equation;
Recall that s(t)=64t-16t^2
s(2) = 64(2) - 16(2)^2
s(2) = 128 - 16(4)
s(2) = 128 - 64
s(2) = 64 ft
Hence the maximum height reached by the ball is 64 feet