A body falling freely from a given height `H` hits an inlclined plane in its path at a height `h`. As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of `h//H`, the body will take the maximum time to reach the ground.

Explanation

In accordance with Fig. 5.48, timet taken by the body to strike the inclined plase
`t_(1)=sqrt((2(H-h))/(g))`
Now, as after impact the velocity of thebody is horizontal, so time taken to reach the ground
`t_(2)=sqrt((2h)/(g))`
So, total time of motion,
`t=t_(1)+t_(2)=sqrt((2)/(g))[sqrt(h)+sqrt((H-h))]`
For t to be maximum (dt/dh)=0
i.e., `(d)/(dh)[sqrt((2)/(g))(h^(1//2)+(H-h)^(1//2)]=0`
or`(1)/(2)h^(-1//2)+(1)/(2)(H-h)^(-1//2)(-1)=0["as"sqrt((2)/(g))cancel=0]`
or `h=H-h, i.e., (h)/(H)=(1)/(2)`