Re: What is the speed of a served tennis ball as it crosses the net?

The serve of a tennis ball is similar in some ways to the pitch of a baseball (see Re: When a pitcher throws a ball, when is the ball at its maximum speed?) except for one very important thing: the tennis ball and air interaction is the turbulent regime.    You might also wish to look at Re: How does tennis relate to physics?

This basically means that because of the rough surface of the the tennis ball during a power serve is traveling at a fast enough speed that the air separating from both sides of the tennis ball becomes turbulent.   At this speed the backspin of the tennis will cause it to rise which is a positive Magnus effect.    This is a the same effect that causes topspin lobs to die and fall straight down or other ball-air effects like the flight of a golf ball.  For more background use  MadSci Search using keywords: air resistance ball.

Anyway, if you assume a flat trajectory what you are implying is that the positive Magnus effect is pretty much cancelling out the drop due to gravity.   If this assumption is true you need to know how the air is also slowing the ball down as it flies.    To pretty good approximation

F_{air drag} = - C * speed²

where C is some constant depending on the cross section of the spherical ball and speed is how fast the ball is moving relative to the air.  This force is negative indicating that is in the opposite direction of the motion of the ball.   This is an assumption!  The air drag force on a ball is approximately proportional to speed squared but can change with different speeds and spin, and the constant C can change as well with fluid dynamic range.

Let me assume that the tennis ball starts at a height of around 9 ft (average ht of a player about 5.5 ft with an arm and raquet above the head of about 3.5 ft).   This gives me an angle of the path the ball travels.   If the ball travels on a flat trajectory then the distance can be computed using trigonometry.

Angle of descent  = tan ^-1 (9/59) = 8.7 degrees.

distance of straight path = 59 ft / cos (8.7 degrees) = 59.7 ft

If we assume this is true (and this a very sketchy assumption) then we can proceed as follows:

Energy of Ball-Air System (ignoring gravity because of positive Magnus Force) = 1/2 mass * speed ² -Fair drag * d

where I have defined F_{air drag} to be negative and mass is the mass of the ball and d is the distance travelled along the flight path.  The term Fair drag * d is the work lost(energy) from

Substituting in my expression for F_{air drag}

Energy of Ball-Air system (ignoring gravity because of positive Magnus Force) = 1/2 mass * speed² + C*speed² * d
 

This energy is conserved at least until the ball collides with the ground.   It has the basic form I would expect: the speed of the ball must slow down the distance d increases.
 
 
 

The ball starts with an initial speed of 120 mph and d = 0 and finishes with a speed of 87 mph and d = 59.7 ft. So this looks like I need to solve two equations with two unknowns.   Presumably I could get a value for mass but it is not necessary. In fact, I could borrow a trick from general relativity and express the mass in a bizarre unit that works for my equation (Joules / (mph)² or (Ft-Lb)/(mph)²) and this allows me to ignore the units while I do the algebra..  As it turns out with this equation, I will be able to solve everything relative to the mass because I don't know it.
 

E_initial = 1/2 mass * 120² + 0 * C

E_final = 1/2 mass * 87² + 87² * 59.7 *C

using conservation of energy (as stated above E_initial = E_final )

7200 * mass = 3784.5 mass +  451869.3 * C
 

C  = 0.0076 mass [units are Joules / (mph)2]

E(d, speed) = 1/2 mass * speed² + 0.0076 mass * speed² * d

[ where E not known but assumed constant, speed in mph, d in feet]

7200 * mass = 1/2 * mass *speed² + 0.0076 mass * 38 * speed²

Notice now that the mass is irrelevant because it cancels out.
 

7200 = (1/2 + 0.289 ) speed²
 

speed² = 7200 / .789 = 9125.5

speed = 95.5 mph
 

So, the speed at the net is about 96 mph.   This seems pretty reasonable, but remember the assumption that went into this: drag coefficient is the same, a flat trajectory means gravity can be ignored, and that energy conservation is simple to apply by considering work done by air drag force over the distance of travel.

Note:  A simple check of the equation by using d = 59 ft gives a speed of 87.1 mph which is perfectly fine with the round-offs I did during calculation.
 

You could try solving this problem with different equations for the force of air drag, but except for the case of speed squared or speed the first power, the solution is difficult to obtain.
 

Sincerely,

Tom "Baseline" Cull