Question

A projectile is fired with a speed of u at an angle $\theta$ above the 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 make its second collision with the field?

Answer 1

Firstly, calculate the time of flight for the projectile then find the distances from the starting point when the projectile will land at its first collision and second collision. After, adding both the positions we can calculate the final answer.
Formula used: $t = \dfrac{{2 \times u\sin \theta }}{g}$

Complete answer:
Given that: initial velocity of the projectile = u and the angle of projection of the projectile with respect to ground = $\theta$.
When the projectile hits the ground for the first time, the velocity remains the same i.e. u. the component of the velocity parallel to ground, $u\cos \theta$ should remain constant. However, the vertical component of the projectile undergoes a change after the collision.

If the coefficient of restitution of the collision between the projectile and the field is e. using the formula of time of flight we know that $t = \dfrac{{2u\sin \theta }}{g}$
Where ‘u’ is the initial velocity, ‘g’ is the acceleration due to gravity and ‘θ’ is the angle of projection. For first collision
${t_1} = \dfrac{{2u\sin \theta }}{g}$
Now position will be given by,
\eqalign{ & {R_1} = \dfrac{{2u\sin \theta }}{g} \times u\cos \theta \cr & \Rightarrow {R_1} = \dfrac{{{u^2}\sin 2\theta }}{g}.......\left( 1 \right) \cr}
As we know, $2\sin \theta \cos \theta = \sin 2\theta$
${R_1}$Is the position of the projectile after the first collision, which is $\dfrac{{{u^2}\sin 2\theta }}{g}$
Now, for the second collision
${t_2} = \dfrac{{2eu\sin \theta }}{g}$
We know that position after second collision is given by
\eqalign{ & {R_2} = \dfrac{{2eu\sin \theta }}{g} \times u\cos \theta \cr & \Rightarrow {R_2} = \dfrac{{e{u^2}\sin 2\theta }}{g}........(2) \cr}
Now, the position of the projectile from the starting point to its final point is given by
$R = {R_1} + {R_2}$
From (1) and (2) we get
\eqalign{ & R = \dfrac{{{u^2}\sin 2\theta }}{g} + \dfrac{{e{u^2}\sin 2\theta }}{g} \cr & \therefore R = \dfrac{{{u^2}\sin 2\theta }}{g}\left( {1 + e} \right) \cr}
This is the final answer.

Note:
The component of velocity parallel to ground, that is $u\cos \theta$ should remain constant. But, $u\sin \theta$ which is the vertical component will change after the collision. As we know the time taken by an object to travel distance is known as time of flight.