A 50.0-g Super Ball traveling at 29.5 m/s bounces off a brick wall and rebounds at 16.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 4.70 ms, what is the magnitude of the average acceleration of the ball during this time interval?

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

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