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?

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

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.