Trajectory of Tennis Ball

Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ below the horizontal. The baseline from which the ball is served is 11.9 m from the net, which is 0.91 m high. What is the angle θ such that the ball just crosses the net? Will the ball land in the service box, which has an outermost service line 6.40 m from the net?
Assumptions:

Assume that the friction force of the ball moving through the air is negligible. After imparting the initial velocity V to the ball, the only force acting on the ball is the gravitational force.

Variables:

At time t, the position of the tennis ball (green) is given by its coordinates x(t) and y(t). Note that positive values of x are to the right, while positive values of y are down. Initially the magnitude of the velocity of the tennis ball is at t=0 is given by V. The velocity vector is inclined at an angle below the horizon. This angle is positive when measured clockwise as indicated in the diagram.

Newton’s Second Law states that the force applied to a body is equal to its mass times its acceleration. Forces, velocities and accelerations are vectors having components (in this case) in the x and y directions. The only force acting on the tennis ball is gravity which is in the vertical direction. That means there is no acceleration in the horizontal direction and the velocity of the tennis ball (in the horizontal direction) remains unchanged from its initial value. If the magnitude of that velocity is V, then the horizontal component of the velocity is given by

(1) $ v_x=V\cos(\theta)$

The time for the ball to strike the plane of the net is the distance to the net (L) divided by the horizontal component of the velocity

(2) $ t_n=\frac{L}{V\cos\theta}$

In the vertical direction the force of gravity acts on the tennis ball, this causes the arc of the trajectory to curve downward. While the initial velocity (downward) of the ball is the downward speed will increase due to gravitational acceleration. Newton’s law for the acceleration in the vertical direction can be written

(3) $ Mg = M\frac{dv_y}{dt}$

Here, the mass of the tennis ball cancels on both sides of the equation yielding the result that the change in velocity is independent of the mass. Integrating equation (3) leads to the result

(4) $ v_y(t)= V\sin(\theta) +gt$

where the first term on the right hand side is the initial vertical velocity. Integrating equation (4) a second time we get the familiar equation for the distance that an object falling in a gravitation field will travel.

(5) $ y(t) = tV\sin(\theta) +\frac{1}{2}gt^2 $

The time at which the ball passes the plane of the net is given by (2). Therefore, the vertical distance of the ball at the time it passes the net is found by using from equation (2) and substituting into (5). If at this time y=H-N (the ball just passes over the net then we get the following equation.

(6) $ H-N = \frac{L}{V\cos(\theta)}V\sin(\theta)+ \frac{L^2g}{2V^2\cos^2(\theta)}$

Just as a check to see if equation (6) is correct, let’s look at the units of each of the three terms. If expressed properly each of the three terms has the units of distance.

The ball hits the court when $y(t_c)=H. From (5) we write:

(7) $H=t_cV\sin(\theta) +\frac{1}{2}gt_c^2 $

Solving for $t_c$

$ t_c=\frac{2[H-V\sin(\theta)]}{g} $

At this time the distance the ball would have traveled is given by multiplying (1) by $t_c$

(8) $ D= Vcos(\theta)t_c $

import sympy as smp
import numpy as np
import matplotlib.pyplot as plt
smp.init_printing()
# Define symbolic variables
g,L,V,H,N,D,tn,tc,theta=smp.symbols('g L V H N D tn tc theta',real=True,positive=True)
g=9.81
N=.91
L=11.9
H=2.5
V=170/3.6
tn=L/(V*smp.cos(theta))
tc=tn*D/L
eq1=H-N-tn*V*smp.sin(theta)-smp.Rational(1,2)*(g*tn**2)
eq2=H-tc*V*smp.sin(theta)-smp.Rational(1,2)*g*tc**2
smp.solve([eq1,eq2],theta,D,tc)

$\displaystyle \left[ \left( 0.106730478959682, \ 17.1947209502893, \ 0.366207329404201\right), \ \left( 1.5968925892839, \ 11.9235420443349, \ -9.67675731360898\right)\right]$

Discussion:

A paper and pencil solution of this problem can be found at https://falkenburg-genealogy.com/notebooks/tennis.pdf

Sympy produces two solutions to this problem. Each solution is a tuple (theta, D, tc) the angle, the distance traveled to hit the court, and the time to hit the court. Notice that the second solution has a negative time tc. Also, the angle theta is larger than pi/2, which means the serve is to the left, not toward the net. Since line 6 requires that the symbols are all positive, I'm not sure why this occurs.Perhaps if lines 12 and 13 were rewritten to be included in the equation list for smp.solve, this would not happen. The first solution (black curve) and the solution in which tc is negative (red curve)are shown below.

For the solution in positive time, the trajectory becomes reaches a peak, passes over the net, and falls short of the foul line at 11.9m.

We can get rid of this anamolous solution by recognizing that solve produces a list of tuples. If we replace line 16 by smp.solve([eq1,eq2],theta,D,tc)[0] (using index 0) we will get the proper solution.

Discussion Since there are 6 unknowns and 6 equations, we find the solution given above in which the initial velocity of the platform is 36.29m/s and the acceleration is 4*g. We will rerun the program in which we comment out # the values for v0,y0,g, and a. this time the solution is given in symbolic form.

We now get two solutions which depend on the relative size of a. Both solutions may not be valid since depending on the value of a tc or u0 may be negative.