Answer
See the detailed answer below.
Work Step by Step
The tennis ball is analogous to a quantum particle in this scenario.
We need to find the tunneling probability for a tennis ball traveling at a given speed when it encounters a potential barrier.
To calculate the energy required to break the strings, we first compute $ U_0 $. This is the kinetic energy of the ball at $ v_{\text{max}} $:
$$
U_0 = \frac{1}{2} m v_{\text{max}}^2
$$
Substitute the known; where the maximum velocity $ v_{\text{max}} = 200 \; {\rm mph} =\bf 89.41 \;\rm m/s$
$$
U_0 = \frac{1}{2} (0.100)(89.41 )^2 = 399.71 \; \text{J}\approx \bf 400\;\rm J
$$
Now we need to find the penetration distance $ \eta $ which is given by
$$
\eta = \frac{\hbar}{\sqrt{2m(U_0 - E)}}
$$
Where $ U_0 = 400\; \text{J} $ which is the potential energy of the barrier, and $ E = \frac{1}{2} mv^2= \frac{1}{2} (0.1)( 53.64)^2=144\;\rm J$ which is the kinetic energy of the tennis ball (we converted 120 mph to m/s).
Substitute the known:
$$
\eta = \frac{(1.05 \times 10^{-34})}{\sqrt{2 ( 0.100) (400- 144) }}=\bf 1.48\times 10^{-35} \;\rm m
$$
The tunneling probability is given by:
$$
P_{\text{tunnel}} = e^{-2w/\eta}
$$
where $ w = 0.002 \; \text{m} $ is the width of the barrier.
Substitute the known:
$$
P_{\text{tunnel}} = e^{-2 (0.002) / (1.48\times 10^{-35} ) } = e^{-2.69 \times 10^{32}}
$$
Since $ e^{-2.69 \times 10^{32}} $ is extremely small, we need to use logarithms to express it. Take the base-10 logarithm:
$$
\log(P_{\text{tunnel}}) = (-2.69 \times 10^{32} ) \log(e)=-1.17 \times 10^{32} $$
Thus:
$$
P_{\text{tunnel}} = \color{red}{\bf 10^{-1.17 \times 10^{32}}}
$$
This is an infinitely small number, so the tunneling probability is essentially zero.
