A stone is dropped from a height h, simultaneously another stone is thrown up from the ground which reaches at a height 4 h, the two stones cross each other after time :-

Question

A stone is dropped from a height h, simultaneously another stone is thrown up from the ground which reaches at a height 4 h, the two stones cross each other after time :- (A) √(h/2 g) (B) √(h/8 g) (C) √8 hg  (D) √2 hg

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/bcd4c6981dcc6683f980ece896f22127-.png" /> If u is the velocity of projection, then 0=u2-2 g(4 h) ∴ u =√8 gh Now y=(1/2) g t2 ...(i) and h - y = ut -(1/2) gt 2 ... (ii) From above equations, we have t =( h / u )=( h /√8 gh )=√( h /8 g )

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

$\frac{h}{2 g}$

$\frac{h}{8 g}$

$8 h g$

$2 h g$

If $u$ is the velocity of projection,
then $0 = u^{2} - 2 g (4 h)$
$∴ u = 8 g h$
Now $y = \frac{1}{2} g t^{2}$ ...(i)
and $h - y = u t - \frac{1}{2} g t^{2}$ ... (ii)
From above equations, we have
$t = \frac{h}{u} = \frac{h}{8 g h} = \frac{h}{8 g}$

Q. A stone is dropped from a height h, simultaneously another stone is thrown up from the ground which reaches at a height 4h, the two stones cross each other after time :-

2gh​​

8gh​​

8hg​

2hg​