The horizontal stream of `H_2O` leaves an opening in the side of a tank. If the opening is h metre above the ground, and the stream hits the ground D metre away and the acceleration due to gravity is g the speed of `H_2O` as it leaves the tank in terms of g, h and D is

Options

A. `D(g/(2h))^(3//2)`
B. `D(g/2h)^2`
C. `D(g/2h)^(-1//2)`
D. `Dsqrt(g/(2h)`

Explanation

Banking angle,
`theta=(tan^-1) (v^2)/(rg)`
`=tan^-1(35^2/(490xx10))
theta=tan^-1(0.25)`