Howard Brody, emeritus professor of physics at the University of Pennsylvania, serves up an answer to this question.

Tennis can be played on almost any surface that can be made smooth enough to provide a consistent bounce of the ball: from acrylic to clay to grass. The interaction of a tennis ball and the surface of a tennis court during the five milliseconds that the two are in contact determines the character of the game for each of the different surfaces. On some surfaces, such as grass, the ball does not bounce very high; on others, like clay, it slows down a great deal after impacting the court. Yet the same laws of physics govern the resulting speed, spin and angle for each type of surface.

There are two dominant forces that act on the ball during those five milliseconds: the normal force (in the vertical direction) and the friction force, which is parallel to the surface. For the bounce of a typical ground stroke—a forehand or backhand shot—the ball impacts the surface at a small angle (about 15 degrees relative to the surface). If the ball could be treated as a point mass—so that it has no real size—then the analysis of the bounce would be straightforward. The ball, however, has properties such as a radius, a moment of inertia, spin, and energy loss when compressed by impact, in addition to mass and friction. The normal force of the surface on the ball is always upward, slowing the ball's downward velocity and propelling it skyward. The friction force is usually backward, reducing the horizontal component of the ball's forward velocity.

Friction may act differently, though, if the ball has excessive topspin, or forward rotation of the ball. In that case, the friction force on the ball is in the same direction as the horizontal component of the ball's velocity. Thus, the ball will have a greater horizontal velocity after the bounce than before the bounce.

How can this possibly happen?

When a ball is spinning at a rate of F revolutions per second, the outside of the ball has a tangential velocity Vt = 2πRF, where R is the radius of the ball. Let us assume the center of the ball is moving in the horizontal direction at Vcm. If the ball has backspin, or underspin, in which the bottom of the ball is moving forward, the bottom of the ball will be moving at Vcm + Vt, in the same direction as the ball is moving. The friction force between the ball and the court surface will be opposite to this, slowing the ball down.

For a ball with topspin, which is the reverse of underspin, the bottom of the ball will be moving at Vcm - Vt. If Vt is greater than Vcm, the bottom of the ball will be moving backward relative to the court surface. This would mean that the friction force between the ball and surface would be forward, which would actually make the ball go faster after the bounce. Therefore, if the topspin is large enough so that 2πRF is greater than Vcm, the horizontal component of the ball's velocity will increase after the bounce.

How fast must the ball be spinning for 2πRF to be greater than Vcm?

For a typical ground stroke with a horizontal velocity of 25 meters per second, a ball of radius 0.033 meter will have to be spinning at greater than 120 revolutions per second.
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