Student Question

A ball is thrown vertically upward from the roof of a 64-foot tall building with a velocity of 96 ft/sec; its height in feet after t seconds is s(t) = 64 + 96 t - 16 t^2. What is the maximum height the ball reaches?

Expert Answers

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The ball is thrown upwards at 96 ft/s from the roof of a building which is 64 foot tall.

The height of the ball after t seconds is given by the function:

s(t) = 64 + 96t - 16t^2

To find the maximum height of the ball, we first find the derivative of the function that defines the height.

s'(t) = 96 - 32t

Equate this to 0,

96 - 32t = 0

=> t = 96/32

=> t = 3

We also see that s''(t) = -32, which is negative, confirming that at t = 3 we have a maximum value.

At time equal to 3 seconds, the height of the ball is s(3) = 64 + 96*3 - 16*9

=> 208

The required maximum height of the ball is 208 feet.

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