A ball 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?

Question

Physics

Jayvion Dennis

28 April, 16:53

A ball 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?

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Answers (1)

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Porter Randall · Answer 1

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