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.

Answer

0.5

Explanation

For motion from A to B
In vertical direction using `s_(y)=u_(y)t+1/2 a_(y)t^(2)`
`u_(y)=0, s_(y)=-(H-h) and a_(y)=-g -(H-h) =1/2 (-g)t_(1)^(2)`
Which give `t_(1)=[(2(H-h))/(g)]^(1//2) ...........(i)`
For motion form B to C, the body starts moving in horizontal direction
`u_(y)=0, s_(y)=-h and a_(y)=-g -h =-1/2 gt_(2)^(2) rArr t_(2) =sqrt((2h)/(g)) .......(ii)`
Total time of fall `T=t_(1)+t_(2) , T=[(2(H-h))/(g)]^(1//2)+[(2h)/(g)]^(1//2)`
For maximum time of flight `(dT)/(dh)=0 rArr (2/g)^(1//2 [(d)/(dh)(H-h)^(1//2) +(d)/(dh) (h)^(1//2)]=0`
`rArr 1/2 (H-h)^(-1//2) (-1) +(1)/(2) h^(-1//2) =0 (1)/sqrt(H-h) =1/sqrth , H-h=h , h=H/2`