Question

A ball is thrown into the air with a velocity of 40 ft/s. Its height, in feet, after t seconds is given by s(t) = 40t - 16t². Find the velocity (in ft/s) of the ball at time t = 1 second.

Answer 1

The velocity of the ball at time t = 1 second is 8 ft/s.

Step-by-step explanation:

To find the velocity of the ball at time t = 1 second, we need to find the derivative of the height function s(t) = 40t - 16t². The derivative of s(t) represents the velocity function v(t). So, v(t) = 40 - 32t. Plugging in t = 1, we get v(1) = 40 - 32(1) = 8 ft/s. Therefore, the velocity of the ball at time t = 1 second is 8 ft/s.