Question

A ball is thrown from the ground with velocity u at
an angle with horizontal. The horizontal range of
the ball on the ground is R. If the coefficient of
restitution is e, then the horizontal range after the
collision is
​

Answer 1

Answer:

u^2 e sin(2@) /g

Explanation:

alright so break vel.

ucos(@)........remains unchanged after collision

usin(@) gets changed to eusin(@)

now for time we study. y directional motion

T=2eusin(@)/g

range = u^2 2cos(@) sin(@) e/g

=u^2esin(2@)/g