A ball is projected on a very long floor. There may be two conditions `(i)` floor is smooth `& (ii)` the collision is elastic
If both the considered then the path of ball is as follows.

Now if collision is inelastic and surface is rough then the path is as follows.

Successive range is decreasing .
Roughness of surface decreases the horizontal component of ball during collision and inelastic nature of collision decreases the vertical component of velocity of ball. In first case both components remain unchanged in magnitude and in second case both the components of the velocity will change.
Let us consider a third case here surface is rough but the collision of ball with floor is elastic. A ball is projected withe speed `u` at an angle `30^(@)` with horizontal and it is known that after collision with the floor its speed becomes `(u)/(sqrt(3))` . Then answer the following questions.
The ratio of maximum height reached by ball in first loop and second loop `((H_(1))/(H_(2)))` is `:`

Options

A. `(1)/(4)`
B. `(1)/(2)`
C. 1
D. `(1)/(sqrt(3))`

Answer

C

Explanation

As the collision is elastic vertical component remains unchanged but the rough floor changes the horizontal component.
`:. U^('2)=((U)/(sqrt(3)))^(2)-(U sin 30^(@))^(2):. U'=(U)/(2sqrt(3))`

Now `tan alpha=((U')/(U sin theta ))^(-1)=sqrt(3) :. alpha=60^(@)`.
As the verticla components remain unchanged therefore the vertical height achieved will remain same.
`:. (H_(1))/(H_(2))=1`