Since sphere is performing pure rolling motion
So here the contact point of sphere with respect to inclined plane is always at rest.
This contact point will not move at any instant and and hence sphere will perform pure rolling.
So here we can say that sphere will have some friction force at the contact point but there is no displacement of that point.
So the work done by this static friction will always ZERO.


So there is no work done by friction on the sphere while it is in pure rolling.
After the collision, the momentum didn't change, so the total momentum in x and y are the same as the initial.
The x component was calculated by subtracting the initial momentum (total) minus the momentum of the first ball after the collision
In the y component, as at the beginning, the total momentum was 0 in this axis, the sum of both the first and struck ball has to be the same in opposite directions. In other words, both have the same magnitude but in opposite directions

This is for both balls after the collision, but one goes in a positive and the other in a negative direction.
Answer:
Explanation:
Given that
The window height is 2m
And the window is 7.5m from the ground
Then the total height of the window from the ground is 7.5+2=9.5m
It takes the ball 0.32sec travelled pass the window.
When the ball get to the window, it has an initial velocity (u') and when it gets to the top of the window it has a final velocity ( v')
Now using the equation of free fall during this window travels
S=ut-½gt² against motion.
S=2, g=9.81, t=0.32sec
Then,
S=u't-½gt²
2=u'×0.32-½×9.81×0.32²
2=0.32u'-0.5023
2+0.5032=0.32u'
Then, 0.32u'=2.5032
u'=2.5032/0.32
u'=7.82m/s
This is the initial velocity as the ball got the the window
Now, let analyse from the window bottom to the ground which is a distance of 7.5m
Using the equation of free fall again
v²=u²-2gH
In this case the final velocity (v) is the velocity when the ball reach the bottom of the window i.e u'=7.82m/s,
While u is the original initial velocity from the throw of the ball
Then,
u'²=u²-2gH
7.82²=u²-2×9.81×7.5
61.146=u²-147.15
61.146+147.15=u²
Then, u²=208.296
So, u=√208.296
u=14.43m/s
The initial velocity of the ball form the throw is 14.43m/s
Hi C is the Answer I believe