Question

A stone dropped from the top of a building of height $h$. It reaches the ground after $t\text{s}$. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity $u$, and they reach the earth surface after $t_1$ and $t_2$ seconds respectively, then

• A. $t=t_1-t_2$
• B. $t=\frac{t_1+t_2}{2}$
• C. $t=\sqrt{t_1{t}_2}$
• D. $t=t_1^2{t}_2^2$

Answer 1

The correct option is C

$t=\sqrt{t_1{t}_2}$

If a stone is dropped from a height $h$ then

$h=\frac{1}{2}g{t}^2$ .....(i)

If a stone is thrown upward with velocity $u$ then

$h=-u{t}_1+\frac{1}{2}g{t}_1^2$ ........(ii)

If a stone is thrown downward with velocity $u$ then

$h=u{t}_2+\frac{1}{2}g{t}_2^2$ ............(iii)

From (i)(ii) and (iii) we get

$-u{t}_1+\frac{1}{2}g{t}_1^2=\frac{1}{2}g{t}^2$ .................(iv)

$u{t}_2+\frac{1}{2}g{t}_2^2=\frac{1}{2}g{t}^2$ ..................(v)

Dividing (iv) and (v) we get

$\therefore \frac{-u{t}_1}{u{t}_2}=\frac{\frac{1}{2}g(t^2-t_1^2)}{\frac{1}{2}g(t^2-t_2^2)}$ or $-\frac{t_1}{t_2}=\frac{t^2-t_1^2}{t^2-t_2^2}$

By solving $t=\sqrt{t_1{t}_2}$