Question

A ball is thrown vertically upward from the ground at a velocity of 125 feet per second. Its distance from the ground after t seconds is given by s(t)=- 16t2 + 125t. How fast is the ball moving 2 seconds after being thrown?

Answer 1

The velocity of the ball at time t is given by the derivative of s(t) with respect to t:

v(t) = s'(t) = -32t + 125

To find the velocity of the ball 2 seconds after being thrown, we can substitute t = 2 into the velocity equation:

v(2) = -32(2) + 125 = 61 feet per second

Therefore, the ball is moving at a velocity of 61 feet per second 2 seconds after being thrown.