Question

A projectile is fired with a speed $u$ at an angle $\theta$ above a horizontal field. The coefficient of restitution of collision between the projectile and the field is $e$. How far from the starting point, does the projectile makes its second collision with the field.

Answer 1

when projectile is projected then its range $R_1=\frac{u^2\sin 2\theta }{g}$

when particle collide with ground then, restitution $e=\frac{v_2-v_1}{u_1-u_2}$

$e=\frac{0-v_2}{-u\sin \theta -0}$

this gives, $v_2=eu\sin \theta$

Now for range , since angle of projection remains same $r_2=\frac{2u_y{u}_x}{g}=e\frac{u^2\sin 2\theta }{g}$

Therefore Range from point of projection will be :$R+R_1+R_2=\frac{u^2\sin 2\theta }{g}(1+e)$