The story of a mercilessly thrown sphere...

Taking the center of the sphere at the origin ,equation of motion of the sphere are: $\sum F_x = Ma_x = -f = \mu_k Mg …………(1)$ $\sum F_y = N - Mg = 0 …………(2)$ From the equations (1) & (2) we get $a = -\mu_k g ………..(3)$ The velocity of the centre of mass at time t is $\upsilon=\upsilon_0 + a_x t.$ or $\upsilon=\upsilon_0 - \mu_k g t ................(4)$ Equation of motion for rotation is $\sum \tau=\mu_k MgR= \Big( \dfrac{2}{5} MR^2\Big) \alpha .................(5)$ or $\alpha =\dfrac{5}{2} \times \dfrac{\mu_k g}{R} ..............(6)$ Then the angular velocity at time t is $\omega=\omega_0 + \alpha t =\dfrac{5}{2} \times \dfrac{\mu_k gt}{R} ...................(7)$ condition for pure rolling is $\displaystyle{ \upsilon = \omega R}$ Now by using (4) & (7) we get $\upsilon_0 - \mu_k g t = \dfrac{5}{2} \times \dfrac{\mu_k gt}{R}$ or $\boxed{ t= \dfrac{2}{7} \times \dfrac{\upsilon_0}{\mu_k g} }$

Conserving angular momentum about any point in the horizontal plane, we get

$mv_{0}R=mvR+\frac{2}{5}mR^{2}\left(\frac{v}{R}\right)^{2}\Rightarrow v=\frac{5}{7}v_{0}$

and

$\frac{5}{7}v_{0}=v_{0}-\mu gt\Rightarrow t=\frac{2v_{0}}{9\mu g}$

OR

we can solve for $t$ from the following equations

$\mu mg=ma$

$\mu mgR=\frac{2}{5}mR^{2}\alpha$

$v_{0}-at=\alpha$

This gives the same answer for $t$

Please be so kind to mention the next time whether the ball is hollow or filled it was certainly a bit confusing

you should give in question that the ball is solid sphere not hollow

FIRST BY USING CONSERVATION OF ANGULAR MOMENTUM ABUT POINT OF CONTACT FIND THE VELOCITY OF CENTER OF SPHERE time is 2/7 times yhe given exp