Abaseball (m = 145 g) approaches a bat horizontally at a speed of 38.9 m/s (87 mi/h) and is hit straight back at a speed of 45.3 m/s (101 mi/h). estimating that the ball is in contact with the bat for a time of 1.00 ms, what is the average force exerted on the ball by the bat? choose the direction of the incoming ball as the positive direction.

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j