A stone is dropped from a height h. Simultane | vedclass

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(A) For the first stone dropped from height $h$,the initial velocity $u_1 = 0$. The distance covered by it in time $t$ is $h_1 = \frac{1}{2}gt^2$.For

Vedclass · Accepted Answer

$\sqrt{\frac{h}{8g}}$

Vedclass · Answer

(A) For the first stone dropped from height $h$,the initial velocity $u_1 = 0$. The distance covered by it in time $t$ is $h_1 = \frac{1}{2}gt^2$.For the second stone thrown upwards from the ground,it reaches a maximum height $H = 4h$. Using the formula $H = \frac{u_2^2}{2g}$,we get $4h = \frac{u_2^2}{2g}$,which implies $u_2^2 = 8gh$ or $u_2 = \sqrt{8gh}$.The distance covered by the second stone in time $t$ is $h_2 = u_2t - \frac{1}{2}gt^2 = \sqrt{8gh}t - \frac{1}{2}gt^2$.Since the stones cross each other,the sum of the distances covered by them must equal the total height $h$,so $h_1 + h_2 = h$.Substituting the expressions: $\frac{1}{2}gt^2 + (\sqrt{8gh}t - \frac{1}{2}gt^2) = h$.This simplifies to $\sqrt{8gh}t = h$.Solving for $t$,we get $t = \frac{h}{\sqrt{8gh}} = \sqrt{\frac{h^2}{8gh}} = \sqrt{\frac{h}{8g}}$.

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

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