# Tennis physics: when does a served ball stop accelerating?

When during a serve does a tennis ball stop accelerating? Is it the moment it leaves contact with the raquet? Or is the ball still accelerating by the time it hits the opponent’s court?

Note that I’m asking about acceleration in the direction the ball is hit—I’m aware that gravity is constantly accelerating everything downwards.

There are two forces on a tennis ball-- the force of impact of the raquet, and gravity. The acceleration from the force of impact is over as soon as the force is over. F = ma. No force, no acceleration. Gravity, OTH is always acting on the ball.

One other thing. Technically, the force of friction is acting to de-accelerate the ball as soon as it starts moving. And when it hits the court, that is an inelastic collision, so some de-acceleration then, too.

After the ball leaves the racket what could be acting on the ball to accelerate it?

In the absence of a force, there is no acceleration, so it stops accelerating when it leaves contact with the racket.

From that moment, the only forces acting on it are gravity (as you mentioned) and drag, which will cause it to decelerate.

The Mythbusters “bullet drop” experiment shows this up very clearly: a bullet fired perfectly horizontally from a gun hits the ground at the same time as a bullet dropped vertically from the same height as the bullet in the barrel. (minus slight corrections due to aerodynamic drag on the bullet, and the curvature of the earth/levelness of the ground over the flightpath of the bullet).

A tennis ball being hit by a racket is subject to the same laws of physics.

Edit: here is the experiment from Mythbusters. Who’d have thought, the laws of physics actually work in practice

Aero drag is acting on it any time its velocity is non-zero. This includes the brief period when it’s in contact with the racket.

At some point prior to ball/racket separation the contact forces will be less than the aero drag forces on the ball.

The ball will begin decelerating a fraction of a second before it separates from the racket.

The tennis ball can probably accelerate again after a bounce, from the energy stored in the ball as it deforms against a surface. But the ball has de-accelerated greatly at that point and is only recovering a small part of the energy lost since leaving the racket.

Topspin and backspin will accelerate the ball in air also.

In the vertical direction, there is a change in sign of velocity, and so a change in acceleration (which is also a vector). At some point, the ball has zero vertical velocity, so it does undergo acceleration right after the bounce (a = dv/dt).

I’m not sure about this. Doesn’t topspin make the ball drop faster and bounce higher rather than accelerate it forward. With backspin doing the opposite?

This is true, so the answer to the OP is when it comes to rest, or stops spinning. And this is only true because we’ve been told to ignore gravity.

(keep in mind that acceleration is a change in velocity, which is a vector of both speed and direction… as other posts have pointed out)

I think you guys are just confusing psychonaut with all these details. Ignoring all the details being discussed, the ball is accelerated only by the racket, and doesn’t continue to speed up as it travels to the opposite court.

Yes, topspin adds to gravitational acceleration and backspin counteracts it.

While not a tennis ball, Fox recently had a pitch tracking system during a baseball game. It showed that a 90 mph baseball is only going 82 or so by the time it reaches the batter. And that’s only about 55 feet or so. My guess is that this change would be more dramatic for a served tennis ball due to its density and distance travelled.

The key to a drop shot is that it won’t bounce forward (much, if at all) precisely because of the backspin.

That force is applied at the instant of contact.

The ball does not accelerate or begin to spin faster during it’s path unless another force e.g. wind acts on it

Yes, but keep in mind that tennis first serves are generally faster than baseball pitches (at comparable levels of play).

Is there some net measure of acceleration (unintentional pun)? It seems like the ball is slowing down the whole time after it leaves the racket.

As acceleration is a vector, you can add the accelerations due to gravity and wind resistance up and get a net acceleration. It would point diagonally downwards and opposite the direction the ball is going, but I’m not sure if it would point more towards the horizontal or the vertical as I don’t know the relative magnitudes of the wind resistance and gravity.

Wikipedia’s article on the Magnus effect shows what’s going on with the ball’s spin. The force imparted by the spin operates perpendicular to the ball’s motion through the air (either side to side or up and down but not forward or back). So any spin doesn’t result in any acceleration towards the receiver.

It’s slowing down in the x-direction, but speeding up in the y-direction. The only positive force acting on it after the initial impact is gravity (until it hits something). That’s assuming the initial impact does not impart terminal velocity to it in the y-direction.