Question

A ball is thrown vertically upward from the top of a 96-ft tower, with an initial velocity of 16 ft/sec. Its position function is s(t) = -16t2 + 16t + 96. What is its velocity in ft/sec when it hits the ground?

Answer 1

Answer:

80 ft/s

Step-by-step explanation:

Given that,

A ball is thrown vertically upward from the top of a 96-ft tower, with an initial velocity of 16 ft/sec. Its position function is given by :

$s(t)=-16t^2 + 16t + 96$ ...(1)

When it hits the ground, s(t) = 0

$-16t^2 + 16t + 96=0\\\\t=3\ s$

At t = 3 s it will hit the fround.

We need to find its velocity when it hits the ground.

Differentiate equation (1) to find its velocity.

$v=\dfrac{ds(t)}{dt}\\\\v=\dfrac{d(-16t^2 + 16t + 96)}{dt}\\\\v=-32t+16$...(2)

Put t = 3 in equation (2)

$v=-32(3)+16\\\\v=-80\ ft/s$

Hence, when it will ground its velocity is 80 ft/s.