## Introduction: Tennis Serve, Its Physics and Math

HOW TO SERVE AT TENNIS, the physical requirements

Throw the ball directly up above your head, up to a foot higher than your racket can reach.

Keep your throwing arm up until the ball starts to descend.

Wait until the ball starts to descend, then start swinging the racket and lowering the throwing arm.

Swinging the racket consists of 3 phases: 1) raising the racket and elbow at the same time until the elbow is the farthest away from your chest as possible and the hand is just above your shoulder with the racket behind you (as if getting ready to throw a football), 2) then raising the racket until it is fully extended, and 3) just before, during, and just after making contact with the ball, skimming the racket along its maximum height (don’t try to go up or down, that’s gravity’s job). Concentrate on the point of contact on the ball.

Imagine that the court is under 8 ½ feet of water and that the ball is floating at time of contact. Skim the ball!

It’s okay to flex the wrist up to the time of contact, but not after making contact with the ball.

LET GRAVITY BRING THE BALL DOWN INTO THE SERVICE AREA.

Serve at 55 to 65 miles per hour (similar to watching a car pass you by, as you stand alongside a highway).

## Step 1: THE MATH, Part 1 of 4

When your racket arm is fully extended, the strings are 8½ feet off the ground. The ball takes .7288 seconds to hit the ground from that height. Here’s the math (**2 = squared, sqrt = square root):

s = gt**2 / 2, where s = distance in meters, t**2 = time in seconds squared, and g is the acceleration caused by gravity on a falling object (9.8 meters per second). Rearranged, t = sqrt( 2s / g).

32 feet per second per second / 9.8 meters per second per second = 8.5 feet / x

x = 8.5 feet * 9.8 mps**2 / 32 fps**2

x = 2.603125 meters

t = sqrt( (2 * 2.603125 m) / 9.8 mps**2) )

t = .7288 seconds

## Step 2: THE MATH, Part 2 of 4

The distance from the serving line to the far side of the serving area is 60 feet (21 + 21 + 18). So in order to cover 60 feet in .7288 seconds, the horizontal speed must be 56.13 miles per hour.

60 feet / .7288 seconds = x feet / 3600 seconds (1 hour, 60 seconds times 60 minutes)

x = 296377.607 feet per hour (56.13 mph, dividing by 5,280 feet per mile)

This is the fastest speed the ball can go and still fall in the serving area. However, the ball slows down due to drag (air resistance) so add 10(?) miles to that.

## Step 3: THE MATH, Part 3 of 4

The distance from the serving line to the net is 39 feet (21 + 18), but the net is 3 feet high (8.5ft – 3ft = 5.5ft).

32 feet per second per second / 9.8 meters per second per second = 5.5 feet / x

x = 5.5 feet * 9.8 mps**2 / 32 fps**2

x = 1.684375 meters

t = sqrt( (2 * 1.684375 m) / 9.8 mps**2)

t = .58630197 seconds

## Step 4: THE MATH, Part 4 of 4

So in order to cover 39 feet in .5863 seconds, the horizontal speed must be 45.3536 miles per hour.

39 feet / .5863 seconds = x feet / 3600 seconds (1 hour, 60 seconds times 60 minutes)

x = 239467.8492 feet per hour (45.3536 mph, dividing by 5,280 feet per mile)

This is the slowest speed the ball can go and still clear the net. However, the ball slows down due to drag (air resistance) so add 10(?) miles to that.

## Step 5: CONCLUSION

Serve straight across at 8.5 feet, at between 66 mph and 55 mph.

## Step 6: OTHER

This method is A SIMPLE SERVE. There are more complicated methods that let you serve faster using the Magnus Effect (for instance, Kick Serve). According to a YouTube video, Samuel Groth, an Australian, made the fastest serve ever with a speed of 163.7 mph (263.4 kph) at the Busan (South Korea) Open 2012 Challenger Event on June 25, 2015.

A graph with a line drawn from the service area line to the top of the net and then on to the strings of the extended racket shows that gravity is needed to bring the ball down. Beyond that, adding aerodynamics will bring the ball down even faster (so the ball can then be hit harder). Without either of these, the ball could be served up to the speed of light in a vacuum and it would never fall into the service area. Also, a fluffy ball will drag more than a smooth one, and thus not travel as far horizontally. See graph.

Opponent’s service line | 21’ | 3’ high net | 21’ | Server’s service line | 18’ | Server’s baseline

21 is to 3 as 60 (21 + 21+ 18) is to 18.57.

Apologies to those not living in Burma, Liberia, and the United States of America for our measurement system.

50 miles = 80.46km; 1 foot (1’) = 30.48 cm.

Also, the original formulas did not import well here. Apologies if editing on the fly caused confusion.

Nice, but if only serving was to know the physics... :)

This is awesome! I love taking math into the real world!