A particle is projected with velocity `2 sqrt(gh)` and at an agnle `60^(@)` to the horizontal so that it just clears two walls of equal height h which are at a distance 2h from each other. The time taken by the particle to travel between these two wall is

Options

A. `2 sqrt((2h)/(g))`
B. `sqrt((h)/(2g))`
C. `2 sqrt((h)/(g))`
D. `sqrt((h)/(g))`

Answer

c

Explanation

Given velocity of particle `= 2 sqrt(gh)`
angle `(theta) = 60^(@)`
distance covered by particle = 2h
Now, `2h = 2 sqrt(gh) cos 60^(@) t`
o, `sqrt(h) = sqrt(g) xx (1)/(2) t`
or `t = 2 sqrt((h)/(g))`