Tunneling probability of a ball over barrier

A particle of energy $E$ meeting a barrier of height $V_0 > E$ and width $a$ can tunnel through with probability

$$T \approx T_0\,e^{-2\alpha a}, \qquad \alpha = \frac{\sqrt{2m(V_0 - E)}}{\hbar}, \qquad T_0 = 16\frac{E}{V_0}\left(1 - \frac{E}{V_0}\right)$$

Given: $m = 0.5\ \text{g} = 5\times10^{-4}$ kg, release height 5 m, barrier height 8 m, width $a = 1.5$ m, $g = 9.81$ m/s², $\hbar = 1.055\times10^{-34}$ J s.

Energies

The ball released from rest at 5 m has total energy (its kinetic energy at ground level):

$$E = mgh_1 = 5\times10^{-4}\times9.81\times5 = 0.02453\ \text{J}$$

To climb an 8 m barrier it would need

$$V_0 = mgh_2 = 5\times10^{-4}\times9.81\times8 = 0.03924\ \text{J}$$ $$V_0 - E = 0.01472\ \text{J}$$

Classically, $E < V_0$, so the ball can never cross.

Decay constant

$$\begin{aligned} \alpha &= \frac{\sqrt{2\times5\times10^{-4}\times0.01472}}{1.055\times10^{-34}} = \frac{3.836\times10^{-3}}{1.055\times10^{-34}} \\ &= 3.64\times10^{31}\ \text{m}^{-1} \end{aligned}$$ $$2\alpha a = 2\times3.64\times10^{31}\times1.5 = 1.09\times10^{32}$$

Transmission probability

$$T_0 = 16\times\frac{0.02453}{0.03924}\left(1 - \frac{0.02453}{0.03924}\right) = 16\times0.625\times0.375 = 3.75$$ $$\begin{aligned} T &\approx 3.75\,e^{-1.09\times10^{32}} \\ &\approx 10^{-4.74\times10^{31}} \approx 0 \end{aligned}$$

Answer: $T \approx e^{-1.09\times10^{32}} \approx 10^{-4.7\times10^{31}}$, which is zero for all practical purposes.

The ball will never be seen tunneling through the barrier, because its mass is huge compared with an electron and the barrier is very wide. Tunneling matters only for very light particles and barriers of nanometre width.