Question

A stone is dropped from a height h. Simultaneously, another stone is thrown up from the ground which reaches a height of $4h$. The two stones cross each other after time

• A. $\sqrt{\frac{h}{2g}}$
• B. $\sqrt{\frac{h}{8g}}$
• C. $\sqrt{8hg}$
• D. $\sqrt{2hg}$

Answer 1

The correct option is B $\sqrt{\frac{h}{8g}}$

Let the initial velocity of stone which is thrown from the ground $u_2$.

At maximum height 4$h$, Velocity $v_2$=0

Using $v^2=u^2-2as$ $\Rightarrow 0=u_2^2-2g\times 4h$ $\Rightarrow u_2=\sqrt{8gh}$ ...1

Let $h_1$ be the distance covered by the stone which is dropped from height $h$,

Let $h_2$ be the distance covered by the stone which is thrown up from the ground.

Let at time $t$ both stone crossed each other.

Using $s=ut+\frac{1}{2}a{t}^2$,

$\Rightarrow -h_1=-\frac{1}{2}g{t}^2$ $\Rightarrow h_1=\frac{1}{2}g{t}^2$ ....2

$\Rightarrow h_2=u_{2}t-\frac{1}{2}g{t}^2$ ....3

Adding $h_1$ and $h_2$ gives the height $h$.

$\therefore$ Adding equations 2 and 3,

$\Rightarrow h=\frac{1}{2}g{t}^2+u_{2}t-\frac{1}{2}g{t}^2$ $\Rightarrow h=u_{2}t$

Using the value of $u_2=\sqrt{8gh}$,

$\Rightarrow h=\sqrt{8gh}t$ $\Rightarrow t=\sqrt{\frac{h}{8g}}$