Physics - W.E.P. Question with Solution | TestHub

$v_C=\sqrt{7gR}$

$mgH=mgR+\frac{1}{2}m(\sqrt{7gR}{)}^2⟹H=R+3.5R=4.5R=\frac{9}{2}R$

$v_D$ : $mg(\frac{9}{2}R)=mg(2R)+\frac{1}{2}m{v}_D^2⟹\frac{5}{2}gR=\frac{1}{2}{v}_D^2⟹v_D=\sqrt{5gR}$

$N_D$ : $N_D+mg=\frac{m{v}_D^2}{R}=\frac{m(5gR)}{R}⟹N_D=5mg-mg=4mg$

$N_C=\frac{m{v}_C^2}{R}=\frac{m(7gR)}{R}=7mg$