Question

If a ball is thrown into the air with a velocity of 36 ft/s, its height (in feet) after t seconds is given by y = 36t − 16t2. Find the velocity when t = 1.

Answer 1

To find the velocity when t = 1, we need to find the derivative of the height function y = 36t - 16t^2 with respect to time t, and then evaluate it at t = 1.

The derivative of y with respect to t is:

dy/dt = 36 - 32t

Evaluating this expression at t = 1, we get:

dy/dt = 36 - 32(1) = 4

Therefore, the velocity of the ball when t = 1 is 4 ft/s.