A particle is projected with velocity `2 sqrt(gh)` so that it just clears two walls of equal height h which are at a distance 2h from each other. Show that the time of passing between the walls is `2 sqrt(h//g)`.

Answer

A

Explanation

`u_(x)^2 + u_(y)^2 = (2(sqrt gh))^2 = 4gh`…….(i)

`v_y = (sqrt(u_(y)^2 -2gh))` ……(ii)
`v_x = u_x` ……(iii)
Now for the projectile ABC, `v_x` and` v_y `are the initial
components of velocity.
`:. 2h = "range" = (2v_x v_y)/g = (2u_x v_y)/g`
or `u_x = (gh)/v_y` .......(iv)
Using Eqs. (ii) and (iv) for rewriting Eq. (i) we have,
`((gh)/(v_y))^2 + (v_(y)^2 + 2gh) = 4gh`
`:. v_(y)^4 - (2gh) v_(y)^2 + g^2h^2 = 0`
`:. v_(y)^2 = (2gh +- (sqrt(4g^2h^2 - 4g^2h^2))) /2`
=gh
`:. v_y = (sqrt(gh))`
Now, `t_(AC) =` time of projectile ABC
`= (2v_y)/g = 2(sqrt h/g)`.