Question

If a ball is thrown into the air with a velocity of 40 ft/s, its height (in feet) after t seconds is given by y = 40t − 16t². Find the velocity when t = 2 seconds.

Answer 1

The velocity of the ball at t = 2 seconds is found by differentiating the height function and substituting t = 2 into the velocity function, giving a result of -24 ft/s.

Step-by-step explanation:

To find the velocity of the ball at t = 2 seconds, we need to take the derivative of the position function y = 40t - 16t², since velocity is the rate of change of position with respect to time. The derivative of y with respect to t is dy/dt = 40 - 32t. Plugging in t = 2, we get dy/dt = 40 - 32(2) = 40 - 64 = -24 ft/s. Therefore, the velocity of the ball at t = 2 seconds is -24 ft/s, which indicates the ball is moving downward.