## Physics (10th Edition)

The range $x$ that the ball reaches is $2m$.
1) Before the tennis ball becomes airborne: The tennis ball has $v_0=0m/s$, $\omega_0=0$ and $h_f-h_0=-1.8m$. Its moment of inertia $I=\frac{2}{3}mr^2$ so $KE_R=\frac{1}{2}I\omega^2=\frac{1}{3}mr^2\omega^2$ According to the principle of energy conservation: $$E_f-E_0=0$$ $$\frac{1}{2}m(v_f^2)+\frac{1}{3}mr^2(\omega_f^2)+mg(h_f-h_0)=0$$ $$\frac{1}{2}m(v_f^2)+\frac{1}{3}mr^2\Big(\frac{v_f^2}{r^2}\Big)+mg(h_f-h_0)=0$$ $$\frac{1}{2}(v_f^2)+\frac{1}{3}(v_f^2)+g(h_f-h_0)=0$$ $$\frac{5}{6}(v_f^2)+g(h_f-h_0)=0$$ $$v_f=\sqrt{-\frac{6g(h_f-h_0)}{5}}=4.6m/s$$ which is the translational speed of the ball just before it becomes airborne. 2) After the ball becomes airborne: The ball's velocity has components $v_x=v\cos35=3.77m/s$ and $v_y=2.64m/s$ On the vertical side, the ball has initial velocity $v_0=2.64m/s$, final velocity $v_f=-2.64m/s$ and acceleration $g=-9.8m/s^2$. The time it is in the air is $$t=\frac{v_f-v_0}{g}=0.539s$$ The range $x$ that the ball reaches is $x=v_xt=2m$