al physics mc

Consider the horizontal component
Horizontal speed of the ball
=10cos30*
=8.66 m/s

Consider the vertical component
Initial vertical speed of the ball
=10sin30*
=5 m/s

The ball will experience a downward acceleration due to gravity.

By the law of conservation of momentum
Loss of P.E. = Gain of K.E.
mgh = 1/2 mv^2
2gh = v^2
Vertical speed of the ball just before reached the ground, v = 14 m/s

As the collision is perfectly elastic
So, vertical speed of the ball just left B = 14 m/s

By s = ut + 1/2 gt^2
0 = 14t - 4.9t^2
t = 2.857 s

The ball is moving with the uniform horizontal speed

Therefore, horizontal distance BC
= horizontal speed X flight time
=8.66 X 2.857
=24.7 m