Two particles A and B are projected with same speed so that the ratio of their maximum heights reach

Question

Two particles A and B are projected with same speed so that the ratio of their maximum heights reached is $3: 1$. If the speed of A is doubled without altering other parameters, the ratio of the horizontal ranges attained by A and B is\

Professor PC Thomas Classes · Answer

$\mathrm{H}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} 	heta_{1}}{2 \mathrm{~g}}$\ $\mathrm{H}_{2}=\frac{\mathrm{u}^{2} \sin ^{2} 	heta_{1}}{2 \mathrm{~g}}$\ Given, $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{3}{1} \Rightarrow \frac{\sin ^{2} 	heta_{1}}{\sin ^{2} 	heta_{2}}=\frac{3}{1}$\ That is $\sin 	heta_{1}=\frac{\sqrt{3}}{2}$ and $\sin 	heta_{2}=\frac{1}{2}$\ $\mathrm{R}_{1}=\frac{2 \mathrm{u}_{1}^{2} \sin 	heta_{1} \cos 	heta_{1}}{\mathrm{~g}}=\frac{2 \mathrm{u}_{1}^{2} \frac{\sqrt{3}}{2} \cdot \frac{1}{2}}{\mathrm{~g}}$\ $\mathrm{R}_{2}=\frac{2 \mathrm{u}_{2}^{2} \frac{1}{2} \cdot \frac{\sqrt{3}}{2}}{\mathrm{~g}} \Rightarrow \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{u}_{1}^{2}}{\mathrm{u}_{2}^{2}}=\frac{4 \mathrm{u}^{2}}{\mathrm{u}^{2}}$\ $	herefore \mathrm{R}_{1} / \mathrm{R}_{2}=4: 1$