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

64 feet

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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

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