First, let us determine the static deformation $y_{static}$ that would be caused by the ball’s weight alone if it were resting on the ground.

When the ball is deformed by a vertical distance $y$ (where $y \ll R$), the contact area is a circle of radius $a$. From geometry (chord theorem or Pythagorean approximation): $$ a^2 = R^2 – (R-y)^2 \approx 2Ry $$ The contact area is $A = \pi a^2 \approx 2\pi R y$.

The upward force exerted by the uniform pressure $p$ is: $$ F_{up} = p \times A = p (2\pi R y) $$ At static equilibrium, this upward force balances the weight $mg$: $$ p (2\pi R y_{static}) = mg \implies y_{static} = \frac{mg}{2\pi R p} $$

The problem states that “half of the elastic potential energy stored during deformation is lost” in each collision.

• Let the ball fall from height $h_n$. Potential energy $PE \approx mgh_n$.
• This energy is converted into elastic potential energy $U_{max}$ at maximum deformation.
• Energy lost = $50\%$ of $U_{max}$.
• Remaining Kinetic Energy for rebound = $0.5 \, U_{max} = 0.5 \, mgh_n$.

Since the rebound energy determines the next height: $$ mgh_{n+1} = 0.5 \, mgh_n \implies h_{n+1} = \frac{h_n}{2} $$ Following this geometric progression, the height after $n$ bounces is: $$ h_n = \frac{h}{2^n} $$

The ball will stop “bouncing” (i.e., it will fail to lose contact with the ground) when the maximum rebound height is insufficient to pull the ball completely out of its static deformation well.

Physically, if the amplitude of the bounce $h_n$ becomes smaller than the static deformation $y_{static}$ required to support its weight, the restoring force at the peak of the “bounce” will still be positive (upward), but the ball will never reach the point $y=0$ (loss of contact). It will merely oscillate around the equilibrium position.

Condition for leaving the ground: $h_n \ge y_{static}$

Substituting our expressions: $$ \frac{h}{2^n} \ge \frac{mg}{2\pi R p} $$

We are given:
Mass $m = 0.2$ kg
Radius $R = 10$ cm $= 0.1$ m
Height $h = 1.0$ m
Pressure $p = 0.2$ atm.
$(1 \text{ atm} \approx 1.013 \times 10^5 \text{ Pa} \implies p \approx 0.2 \times 10^5 \text{ Pa} = 2 \times 10^4 \text{ Pa})$

Calculate the static deformation term ($y_{static}$): $$ y_{static} = \frac{0.2 \times 9.8}{2 \times \pi \times 0.1 \times (0.2 \times 1.013 \times 10^5)} $$ Approximating $g \approx 10$ and $\pi^2 \approx 10$ for quick check, or strictly: $$ \text{RHS} = \frac{1.96}{2 \times 3.1415 \times 0.1 \times 20260} \approx \frac{1.96}{12730} \approx 1.54 \times 10^{-4} \text{ m} $$

Now solve for $n$: $$ \frac{1.0}{2^n} \ge 1.54 \times 10^{-4} $$ $$ 2^n \le \frac{1}{1.54 \times 10^{-4}} \approx 6493 $$

Check powers of 2:
$2^{12} = 4096$
$2^{13} = 8192$

Since $4096 < 6493 < 8192$, the ball can bounce for $n=12$ times. At $n=13$, the height is too small to leave the ground.