Water is filled in a container upto height H as shown in the figure . A small hole is board on the surface of a container at the depth h from the surface of water . What will be the distance of a point along the horizontal where the jet of the water strikes the ground ? For which value of h will this distance be maximum ? Also find this maximum distance.

Explanation

The velocity of water coming out of the hole in horizontal direction is
`v=sqrt(2gh)` …(1)
As the acceleration of water coming out of the hole is downward , it moves with constant velocity in horizontal direction , while it moves with constant acceleration downward . (It is like projectile motion.) From the equations of motion , displacement in downward direction ,
`H-h=(1)/(2)g t^(2)` ....(2)
where t time taken by the water to fall on the ground .
The distance travelled along horizontal, x=vt ...(3)
substituting the values of v and t from equations (1) and (2) in equation (3),
`x=sqrt(2gh)[(2(H-h))/(g)]^((1)/(2))=(4hH-4h^(2))^((1)/(2))`
`=[H^(2)-H^(2)+4hH-4h^(2)]^((1)/(2))`
`x=[H^(2)-(H-2h)^(2)]^((1)/(2))` ...(4)
Equation shows that when `H-2h=0,` then x will be maximum.
`thereforeH-2h=0`
`thereforeh=(H)/(2)`
Putting the value of h in equation (4),
`x=[H^(2)-H-2(((H)/(2))^(x))^(2)]^((1)/(2))`
`thereforex=H`