## Physics (10th Edition)

$v_f=1.3m/s$
The bowling ball has $v_0=3.5m/s$ and $h_f-h_0=0.76m$. It has $I=\frac{2}{5}mr^2$ so $KE_R=\frac{1}{2}I\omega^2=\frac{1}{5}mr^2\omega^2$ According to the principle of energy conservation: $$E_f-E_0=0$$ $$\frac{1}{2}m(v_f^2-v_0^2)+\frac{1}{5}mr^2(\omega_f^2-\omega_0^2)+mg(h_f-h_0)=0$$ $$\frac{1}{2}m(v_f^2-v_0^2)+\frac{1}{5}mr^2\Big(\frac{v_f^2}{r^2}-\frac{v_0^2}{r^2}\Big)+mg(h_f-h_0)=0$$ $$\frac{1}{2}(v_f^2-v_0^2)+\frac{1}{5}(v_f^2-v_0^2)+g(h_f-h_0)=0$$ $$\frac{7}{10}(v_f^2-v_0^2)+g(h_f-h_0)=0$$ $$v^2_f-v_0^2=-\frac{10g(h_f-h_0)}{7}=-10.64$$ $$v_f=\sqrt{v_0^2-10.64}=1.3m/s$$