Question

From the top of tower, a stone is thrown up. It reaches the ground in $t_1$ s. A second stone thrown down with the same speed reaches the ground in $t_2$ s. A third stone released from rest reaches the ground in $t_3$ s. Then

• A. $t_{3} = \frac{\left(t_{1}+t_{2}\right)}{2}$
• B. $t_{3} = \sqrt{t_{1}t_{2}}$
• C. $\frac{1}{t_{3}} = \frac{1}{t_{1}} + \frac{1}{t_{2}}$
• D. $t^{2}_{3} = t^{2}_{2} - t^{2}_{1}$

Answer 1

Answer: (B)

When stone is thrown vertically upwards from the top of tower of height h, then
$h = ut_{1} + \frac{1}{2}gt^{2}_{1}\quad...\left(i\right)$
When stone is thrown vertically downwards from the top of tower, then
$h = ut_{2} + \frac{1}{2}gt^{2}_{2}\quad ...\left(ii\right)$
When stone is released from the top of tower, then
$h = \frac{1}{2}gt^{2}_{3}\quad ...\left(iii\right)$
From equation $\left(i\right)$, we get
$\frac{h}{t_{1}} = -u +\frac{1}{2}gt_{1}\quad ...\left(iv\right)$
From equation $\left(ii\right)$, we get
$\frac{h}{t_{2}} = -u +\frac{1}{2}gt_{2}\quad ...\left(v\right)$
Adding equations $\left(iv\right)$ and $\left(v\right)$. we get
$h\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}\right) = \frac{1}{2}g\left(t_{1}+t_{2}\right)$ or $h = \frac{1}{2}g\,t_{1}t_{2}$
Putting the value in equation $\left(iii\right)$, we get
$t_{3} = \sqrt{t_{1}t_{2}}$