Oct 10, 2014

After t/2 seconds it has travelled a quarter of the total distance. This means that if the height of the tower is $h$ then it has fallen a distance of $\frac{h}{4}$ so is at a distance of $\frac{3 h}{4}$ above the ground.

#### Explanation:

${h}_{t} = \frac{1}{2} g . {t}^{2}$

Where ${h}_{t}$ is the total height and g is the acceleration due to gravity.

After $\frac{t}{2}$ seconds :

${h}_{\frac{t}{2}} = \frac{1}{2} g {\left(\frac{t}{2}\right)}^{2}$

$= g \frac{{t}^{2}}{8}$

So ${h}_{t} = 4 {h}_{t \frac{1}{2}}$

So you can see that after half the time it has travelled a quarter of the distance.

This means that if the height of the tower is $h$ then it has fallen a distance of $\frac{h}{4}$ so is at a distance of $\frac{3 h}{4}$ above the ground.