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) = −16t 2 + 32t+ 240 feet above the ground. What is the velocity of the ball as it hits the ground?

Answer 1

V = (-32t +32) ft/s

The velocity of the bowling ball after t seconds is the time derivative of the position function s(t). To obtain the velocity from s(t) we simply differentiate s(t) with respect to t. That is

V = ds(t)/dt = -16×2t + 32×1 = -32t + 32.

When differentiating, you multiply the coefficient of each term in the equation with the power of the variable and then reduce the power by 1 just like above.

Step-by-step explanation: