Rolling Projectile!

Consider a sphere of mass $'M'$ , and radius $'R'$ colliding with the ground, and rotating with the angular velocity $\omega$ clockwise, as shown below.

Now, consider a vertical impulse ${J}_{v}$ given to the ball in the upward direction by the ground, and a horizontal impulse ${J}_{H}$ along the horizontal direction, but opposite to the angular velocity. The vertical impulse is responsible for the change in the sphere's momentum along the vertical direction whereas the the horizontal impulse is what brings about a change in the sphere's horizontal linear momentum, and it's angular momentum.

Let the height from which the centre of mass falls is $h$ and it's vertical velocity at the instant of contact is ${u}_{y}$ . By principle of energy conservation, we can say that:

$\frac {1} {2} M{{u}_{y}}^{2} = Mgh$

${u}_{y} = \sqrt {2gh}$

Let the final velocity of the sphere in the vertical direction is given by ${v}_{y}$ . This ${v}_{y}$ is what will provide the vertical velocity for the projectile motion of the sphere. Now, by impulse momentum relation, we can easily infer that the magnitude of impulse is given by:

${J}_{v} = M({u}_{y} - {v}_{y})$

Also, let the body's velocity in the horizontal direction be ${v}_{x}$ and it's final angular velocity is ${\omega} {o}$ . Now, for the horizontal direction, the horizontal impulse can be related with change in the sphere's linear momentum and it's angular momentum by:

${J}_{H} = M{v}_{x}$

${J}_{H} R = \frac {2}{5} M{R}^{2} (\omega-{\omega}_{o})$

where $\frac {2}{5} M{R}^{2}$ represents the sphere's moment of inertia about it's centre of mass. The only remaining thing to do now, is to relate ${J}_{v}$ and ${J}_{H}$ , which can be done by considering the coefficient of friction, $\mu$ i.e. ${J}_{H}= \mu {J}_{v}$ .

Now, for our case, $h=20m$ , ${v}_{y} = 4m/s$ , $\omega=7rad/s$ and $R=2m$ .

Plugging in these values, we get:

${J}_{v} = 16m$

And by conserving angular momentum:

$\mu{J}_{v} (2) = \frac {2}{5} M{(2)}^{2} (7-{\omega}_{o})$ $\Rightarrow 7 -{\omega}_{o} = \frac {5\mu{J}_{v}} {4M}$ $\Rightarrow {\omega}_{o}=7-\frac {5\mu{J}_{v}} {4M}$

Since, the question states the condition for pure rolling, hence, we can say that ${v}_{x} = R {\omega}_{o}$ , or:

${v}_{x} = 2 (7-\frac {5\mu{J}_{v}} {4M})$ $\Rightarrow \frac {\mu {J}_{v}}{M} = 14-\frac {5\mu{J}_{v}} {2}$ $\Rightarrow \frac {7\mu {J}_{v}} {2M} = 14$ $\Rightarrow \mu {J}_{v} = 4M$

But, we already know that $\mu {J}_{v} = M {v}_{x}$ , which finally gives us ${v}_{x} = 4m/s$

Now, finally, the range of the projectile can be given by

$\frac {2 {v}_{y} {v}_{x}} {g} = 3.2 m$