Animated Solution for Physics - Kinematics: A ball is dropped from the top of a $100\text{ m}$ high tower on a planet. In the last $\frac{1}{2}\text{ s}$ before hitting the ground, it covers a distance of $19\text{ m}$. Acceleration due to gravity (in ${\text{ms}}^{-2}$) near the surface on that planet is ........... .

Imagine standing at the top of a towering $100\text{ m}$ structure on an unknown planet. You drop a ball, meaning its initial velocity is exactly zero ($u=0$). The problem gives us a fascinating piece of information: in the final $0.5\text{ s}$ before it smashes into the ground, the ball covers exactly $19\text{ m}$.

This implies something crucial about the journey before that final half-second. If the total height is $100\text{ m}$ and the last stretch is $19\text{ m}$, the ball must have fallen $100-19=81\text{ m}$ before the clock started ticking on that final $0.5\text{ s}$.

To unravel the mystery of this planet's gravity, we turn to the second equation of motion:

Since the ball is dropped from rest, the initial velocity $u$ is zero. This beautifully simplifies our equation to:

Let's break the fall into two distinct phases to avoid messy algebra.

Phase 1: The First 81 Meters Suppose it takes time $t_1$ for the ball to fall the first $81\text{ m}$. Plugging this into our simplified equation:

Rearranging to solve for $t_1$, we get:

Phase 2: The Total 100-Meter Fall Now, consider the entire $100\text{ m}$ drop. The total time taken is the time for the first $81\text{ m}$ plus the final $0.5\text{ s}$, which is $t_1+0.5$. Using our equation again:

Solving for the total time term gives:

We now have two elegant expressions. Let's substitute the value of $t_1$ from the first phase into the equation for the total fall:

Now, it's just a matter of simple algebra. Move the terms with $\sqrt{a}$ to one side:

Cross-multiplying yields:

Squaring both sides, we reveal the hidden gravity of the planet: