Aball is thrown up into the air with an initial velocity of 64 ft/sec. the ball is 4 ft above the ground when it is released. its height, in feet, at time t seconds is given by: h(t) = −16t^2 + 64t + 4. what is the time when it reaches its maximum height?

2 seconds and maximum height reached is 68 ft

Explanation:

h (t) = -16t^2 +64t +4

differentiate with respect to t

h'(t) = -32t +64

for maximum height velocity which is h'(t) must be zero

-32t +64 = 0

-32t = -64

t = 2 s

hence height is

h(2) = -16(2)^2 +64(2) +4

h(2) = -64+128+4

h(2) = 68 feet

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