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)`.

Options

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

Answer

C

Explanation

2h= ut
`u_(x)= (2h)/(Delta t) [because u= u_(x), t= Delta f]` …(i)
`h= u_(y) t- (1)/(2) g t^(2)`

`g t^(2)- 2u_(y ) t+2h=0`
`t_(1)= (2u_(y)+ sqrt(4u_(y)^(2)-8gh))/(2g)`
`t_(2)= (2u_(y)- sqrt(4u_(y)^(2)-8gh))/(2g)`
`Delta t= t_(1)-t_(2)= (sqrt(4u_(y)^(2)-8gh))/(g)`
`u_(y)^(2)= (g^(2) (Delta t)^(2))/(4)+ 2gh`
`u_(x)^(2)+ u_(y)^(2)= u^(2)=(2 sqrt(gh))^(2)`
`(4h^(2))/((Delta t)^(2)) + (g^(2) (Delta t)^(2))/(4) + 2gh= 4gh`
`(g^(2))/(4) (Delta t)^(4)- 2gh (Delta t)^(2)+ 4h^(2)= 0`
`(Delta t)^(2)= (2gh +-sqrt(4g^(2)h^(2)-4g^(2)h^(2)))/(g^(2)//2)= (4h)/(g)`
`Delta t=2 sqrt((h)/(g))`