Question

A ball is dropped from a tower. In the last second of its motion it travels a distance of $15m$. Find the height of the tower. [take $g=10m/se{c}^2]$.

Answer 1

Time of flight $=\sqrt{\frac{2H}{g}}s$

Distance traveled in last second =H-distance traveled in $=(\sqrt{\frac{2H}{g}}-1)s$

Distance traveld in $(\sqrt{\frac{2H}{g}}-1)s$=$\frac{1}{2}g(\sqrt{\frac{2H}{g}}-1{)}^2$

$=\frac{g}{2}[\frac{2H}{g}+1-2\sqrt{\frac{2H}{g}}]$

$=H+\frac{g}{2}-g\sqrt{\frac{2H}{g}}$

$=H+\frac{g}{2}-g\sqrt{2H}g$

$15=H-(H+\frac{g}{2}-\sqrt{2gH}$

$=\sqrt{2gH}-\frac{g}{2}=\sqrt{2gH}-5$

$\sqrt{2gH}=20$

$2\times 10\times H=400$

$H=20m$