Question

A ball is dropped from the top of a tower of height h. It covers a distance of $\frac{h}{2}$ in the last second of its motion. How long does the ball remain in air?

Answer 1

For $1^{st}\frac{h}{2}$ fall

$\frac{h}{2}=0t_1+\frac{1}{2}g{t}_1^2$$[$ from $S=\mu t+\frac{1}{2}g{t}^2]$

$\frac{h}{2}=\frac{1}{2}g{t}_1^2\Rightarrow \frac{h}{g}=t_1^2....\left(1\right)$

Also we know

$V=0+g{t}_1\Rightarrow V=g{t}_1$

For $2^{n}d\frac{h}{2}$ fall

$\frac{h}{2}=\left(g{t}_1\right)t_2+\frac{1}{2}g{t}_2^2\frac{1}{2}.\frac{h}{g}=t_1{t}_2+\frac{t_2^2}{2}\Rightarrow \frac{t_1^2}{2}=t_1{t}_2+\frac{t_2^2}{2}\left[from\right(1\left)\right]$

Given that $t_2=1$

$\Rightarrow t_1^2=2t_1+1\Rightarrow t_1^2-2t_1-1=0\Rightarrow t_1=\frac{2\pm \sqrt{4+4}}{2}=\frac{2\pm 2\sqrt{2}}{2}{t}_1=1+\sqrt{2}$

$\therefore$ total time for which ball remains in air.

$t_1+t_2=1+\sqrt{2}+1=2+\sqrt{2}seconds$