Question

A bowling ball is launched off the top of a 240-foot tall building. The height of the bowling ball above the ground t seconds after being launched is s(t)=−16t2+32t+240s(t)=−16t2+32t+240 feet above the ground. What is the velocity of the ball as it hits the ground?

Answer 1

When the ball hits the ground, its velocity is -128 ft/s.

Step-by-step explanation:

Hi there!

First, let's find the time it takes the ball to reach the ground (the value of t for which s(t) = 0):

s(t) = -16t² + 32t + 240

0 = -16t² + 32t + 240

Solving the quadratic equation with the quadratic formula:

t = 5.0 s (the other solution of the equation is rejected because it is negative).

Now, we have to find the velocity of the ball at t = 5.0 s.

The velocity of the ball is the change of height over time (the derivative of s(t)):

v = ds/dt = s'(t) = -32t + 32

at t = 5.0 s:

s'(5.0) = -32(5.0) + 32 = -128 ft/s

When the ball hits the ground, its velocity is -128 ft/s.