Bounce it like Beckham

Let the vertical component of impulse on the ball during collision be $J$ . The instantaneous velocity of the ball after collision would then have a vertical component of $\frac{J}{m}$ . The angular velocity of the ball after collision would be $\omega-\frac{Jr}{\frac{2}{5}mr^2}$ where $\omega$ is the initial angular velocity of the ball. Since friction between the ball and the wall is very high, we can assume that the ball appears to be rolling without slipping at the instant after collision. The point on the ball touching the wall would then have a vertical component of $0$ . Hence, equating the linear and rotational velocities we see that $\frac{J}{m} = (\omega-\frac{Jr}{\frac{2}{5}mr^2})r$ . Rearranging the above equation, we see that the vertical velocity of the ball after collision is $v_y = \frac{J}{m} = \frac{2}{7}\omega r$ . Since collision in the horizontal direction is perfectly elastic, it is clear that the horizontal velocity is the same as the initial velocity of the ball. $v_x = \omega r$ Hence, the angle that the ball makes with the ground would be $\theta = \tan^{-1} \frac{v_y}{v_x} = \tan^{-1} \frac{2}{7} = 15.95^\circ$

Initially ball is rolling with speed v , so initially angular momentum of ball $= (2/5) Mr^2ω = (2/5)Mvr$ ( as for rolling ω = v/r). Now conserving angular momentum on the wall we get 2/5 Mvr = 2/5 Mv'r (due to rolling) + Mv'r (due to translation) hence v' = 2/7 v. Now the ball has two velocities in upward direction ( 2/7v) and in backward direction (v) ( due to elastic collision with wall) therefore tanx = 2/7 where x is the angle made with ground. Therefore x = 16 degrees (approx.)

Let $v$ and $w$ be the initial velocity and angular velocity of the ball, respectively.

In the beginning of the collision process, since the ball already has some angular velocity but no upward velocity (velocity along the wall) it will slip against the wall. However, there is a impulse on the ball due to the friction force with the wall. If $k$ is too small, the ball will slip until the end of the collision process. However, if $k$ is large enough (detailed calculations show that $k>1/7$ is sufficient), the ball will slip at first, and then roll without slipping by the end of the collision. In this problem, $k>>1$ so the ball will roll without slipping at the end.

If the total impulse exerted upward on the ball at the contact point between it and the wall is X, then the final center of mass velocity in the y-direction will be $v'=X/m$ . If we consider the change in angular momentum due to this impulse, and know the moment of inertia for a solid sphere, $I=2mr^2/5$ one can solve for the final angular velocity $w'=w-5X/2mr$ . Since at the beginning and end everything is rolling without slipping,we know that $v=wr$ (initially on the ground), $v'=w'r$ (finally on the wall before it bounces off) $\Rightarrow v'=2v/7$ .

So the jumping angle $c$ is just $tan(c)=v'/v \Rightarrow c=15.95^o$