Power serve

The force acting on the ball is $F=\frac{1}{2}c_f \rho A v^2$ , where $c_f = 0.47$ is the drag coefficient for a sphere, $\rho$ is the density of the air, $A$ is the cross section of the circular area facing the wind and $v$ is the velocity. The equation of motion for the ball is

$\frac{d v}{d t}= -\frac{c_f\rho A}{2 m}v^2=-k v^2$ ,

where $m$ is the mass, $v$ is the velocity and $k=\frac{c_f\rho A}{2 m} = 0.0166m^{-1}$ . Solving this nonlinear differential equation is not trivial, but it becomes much simpler if we introduce the position $x$ as a variable, instead of the time $t$ . We use $\frac{d v}{d t}=\frac{d x}{d t}\frac{d v}{d x}=v\frac{d v}{d x}$ and we get

$v\frac{d v}{d x}= -k v^2$ or $\frac{d v}{d x}= -k v$

At $x=0$ the velocity is $v=v_0$ . With this initial condition the solution is $v=v_0 e^{-kx}$ or $v/v_0 = e^{-kx}$ . We are interested in the quantity $(v_0-v)/v_0= 1- e^{-kx}= 0.259$ . The loss of velocity is about 25%.

Bonus: Initially the drag force on the ball is $F=\frac{1}{2}c_f \rho A v^2=2.9N$ . The weight is $mg=0.56N$ that is more than 5 times less than the drag force. By the time the ball reaches the ground the drag force is only about 2 times larger than the weight.

Here is a link to a serious experimental study . According to the study the drag coefficient is actually a bit larger, $c_f=0.5$ , probably due to the fuzziness of the ball.