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

P.E. of the stone projected vertically is, P.E. = mgh But $\mathrm{h}=\frac{{\mathrm{v}}^2}{2\mathrm{g}}$ $\begin{array}{rl}\therefore \text{ P. }{\mathrm{E}}_1& =\mathrm{mg}\left(\frac{{\mathrm{v}}^2}{2\mathrm{g}}\right)\\ & =\frac{{\mathrm{mv}}^2}{2}...(i)\end{array}$ For the second stone thrown at an angle $\theta$ to the horizontal, $\begin{array}{ll}& \mathrm{h}=\frac{{\mathrm{v}}^2{\sin }^2\theta }{2\mathrm{g}}=\frac{{\mathrm{v}}^2{\sin }^{2}30^{\circ }}{2\mathrm{g}}=\frac{{\mathrm{v}}^2}{8\mathrm{g}}\\ \therefore & \text{ P.E. }2=\mathrm{mg}\left(\frac{{\mathrm{v}}^2}{8\mathrm{g}}\right)=\frac{{\mathrm{mv}}^2}{8}...(ii)\end{array}$ Dividing equation (i) by equation (ii) - $-\frac{P\cdot E_1}{P\cdot E_2}=\frac{\left(\frac{m{v}^2}{2}\right)}{\left(\frac{m{v}^2}{8}\right)}=4:1$