A cubical tank carrying a non viscous liquid (water, say) moves down a smooth inclined plane when we open the hole made at the mid-point `P` of the face of the tank, it strikes the point `Q` of the inclined plane. Find the of the tank at the time of opening the valve.

Answer

`sqrt(ghcostheta)`

Explanation

`v_(rel)=sqrt(2g_(eff)h/2)-sqrt(g_(eff)h)=sqrt(ghcostheta)`
The actual velocity of the liquid `v=(v_(r)-v_(0))`

Then the time of flight `T=sqrt((2(h/2))/(gcostheta))=sqrt(h/(gcostheta))`
`h=vt=1/2(gcostheta)T^(2)` putting `T`,and `v`, we have
`v_(0)=h(1+(tantheta)/2)sqrt((gcostheta)/h)+sqrt(ghcostheta)`