A ball is released from the top of a tower of height $h$ meters. It takes $T$ seconds to reach the ground. What is the position of the ball at $T/3$ seconds? A). $\frac{8h}{9}$ meters from the ground B). $\frac{7h}{9}$ meters from the ground C). $\frac{h}{9}$ meters from the ground D). $\frac{17h}{18}$ meters from the ground

The correct answer is $\frac{8h}{9}$ meters from the ground. **Solution:** Using $s = ut + \frac{1}{2}gt^2$: For the full drop: $h = 0 \times T + \frac{1}{2}gT^2 \Rightarrow h = \frac{1}{2}gT^2$ The vertical distance fallen from the top in time $T/3$ is: $h' = \frac{1}{2}g\left(\frac{T}{3}\right)^2 = \frac{1}{2}g \frac{T^2}{9} = \frac{h}{9}$ Therefore, the position (height) of the ball *from the ground* is: $h - \frac{h}{9} = \frac{8h}{9}$

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