A stone dropped from a building of height $h$ reaches the earth after $t$ seconds. From the same building, if

Question

A stone dropped from a building of height $h$ reaches the earth after $t$ seconds. 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:

Accepted Answer

1. **Analyze the Dropped Stone:** For a stone dropped from rest ($u=0$), the height $h$ is given by $h = \frac{1}{2}gt^2$. Thus, $t = \sqrt{\frac{2h}{g}}$.
2. **Analyze the Stone Thrown Upwards:** Initial velocity is $+u$ (upward), displacement is $-h$ (downward). Using $s = ut + \frac{1}{2}at^2$:
   $$ -h = u t_1 - \frac{1}{2}g t_1^2 \implies h = -u t_1 + \frac{1}{2}g t_1^2 \quad 	ext{...(i)} $$
3. **Analyze the Stone Thrown Downwards:** Initial velocity is $-u$ (downward), displacement is $-h$. Using the same equation:
   $$ -h = -u t_2 - \frac{1}{2}g t_2^2 \implies h = u t_2 + \frac{1}{2}g t_2^2 \quad 	ext{...(ii)} $$
4. **Solve for Relationship:** From (i), $u = \frac{1}{2}g t_1 - \frac{h}{t_1}$. From (ii), $u = \frac{h}{t_2} - \frac{1}{2}g t_2$. Equating $u$:
   $$ \frac{1}{2}g t_1 - \frac{h}{t_1} = \frac{h}{t_2} - \frac{1}{2}g t_2 $$
   $$ \frac{1}{2}g (t_1 + t_2) = h \left( \frac{1}{t_1} + \frac{1}{t_2} ight) = h \frac{t_1 + t_2}{t_1 t_2} $$
   $$ h = \frac{1}{2}g t_1 t_2 $$
   Comparing this with $h = \frac{1}{2}gt^2$, we get $t^2 = t_1 t_2$ or $t = \sqrt{t_1 t_2}$ .

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