Question

A stone is dropped from a height h . Simultaneously another stone is thrown up from the ground with such a velocity that it can reach a height 4h . Find the time when two stone cross each other .


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Answer 1

Explanation:

Your answer refer in Attachment

Thank you

Answer 2

SOLUTION

u is the velocity with which second stone is projected.

${v}^{2} - {u}^{2} = 2as$

At heighest point v= 0

s= 4h

a= -g

0- u^2 = -2g4h

u= √8gh

Let t is the time when they cross each other.

In t time distance travel by first stone is:

$x = ut + 0.5g {t}^{2} \\ = > x = 0 + 0.5 \times 10 {t}^{2} \\ = > x = 5 {t}^{2}$

Distance travelled by second stone is:

$h - x = ut - 5 {t}^{2} \\ = > h - 5 {t}^{2} = \sqrt{8ght} - 5 {t}^{2} \\ = > h = t \sqrt{8gh} \\ \ = > t = \sqrt{ \frac{h}{8g} }$

Hope it helps ☺️