Physics - Center of Mass Question with Solution | TestHub

The block goes$P\to C\to B$ on the ground, climbs the smooth curved track to its top A (height 2R) and flies off horizontally, landing back at P (range 2R). Projectile:$2R=\frac{1}{2}g{t}^2\Rightarrow t=\sqrt{\frac{4R}{g}}=\sqrt{6}$ s;$u=\frac{2R}{t}=\frac{30}{\sqrt{6}}$, so$u^2=150$. Energy:$\frac{1}{2}{V}_0^2=\mu gR+g(2R)+\frac{u^2}{2}=75+300+75=450$.$V_0=30\mathrm{m}/\mathrm{s}$, so$\frac{V_0}{6}=5$. Answer: 5

Answer: 5