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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

The distance covered by a stone dropped from the top of a building in the last second of its motion is $0.36$ times the total distance travelled by it. The height of the building is (acceleration due to gravity $g = 9.8 \ m/s^2$). (in $m$)

$A$ stone is allowed to fall from the top of a tower $100 \ m$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $25 \ m/s$. The two stones will meet after ......... $s$.

$A$ particle is thrown vertically upwards with a velocity of $4\,ms^{-1}$. The ratio of its accelerations after $1\,s$ and $2\,s$ of its motion is

$A$ ball is thrown up vertically with a certain velocity so that it reaches a maximum height $h$. Find the ratio of the times at which it is at height $\frac{h}{3}$ while going up and coming down respectively.

Vedclass Test Series

Exam Paper Generator

Online Exam Module