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.

Options

A. `(1)/(3)`
B. `(1)/(2)`
C. `(2)/(5)`
D. `(2)/(3)`

Answer

B

Explanation

PQR is an inclined plane. The body falls under gravity vertically from A to B till it strikes the inclined plane. At B its velocity is horizontal. From B to C, the body follows a parabolic path. At , C it reaches the ground. Let the time taken to travel from B to `C=t_(2)`
`:'S=ut+(1)/(2)g""t^(2)`
`:.(H-h)=(0xxt_(2))+(1)/(2)g""t_(1)^(2)`
or `t_(1)=sqrt((2(H-h))/(g))`
Similarly `t_(2)=sqrt((2h)/(g))`
`:."Total time",t=t_(1)+t_(2)=sqrt((2h)/(g)][sqrt(H-h))+sqrt(h)]`
`:.(dt)/(dh)=sqrt((2)/(g))[(-1)/(2sqrt(H-h))+(1)/(2sqrt(h))]`
For t to be maximum, `(dt)/(dh)=0`
`sqrt((2)/(g))[(-1)/(2sqrt(H-h))+(1)/(2sqrt(h))]`
or `H-h=himpliesH=2himplies(h)/(H)=(1)/(2):.(h)/(H)=(1)/(2)`