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?

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.