Two bodies are thrown up at angles of 45^o an | vedclass

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Question

$\mathrm{H}_{\max }=\frac{\mathrm{u}^{2} \sin ^{2} 	heta}{2 \mathrm{g}}$

According to problem

$\frac{\mathrm{u}_{1}^{2} \sin ^{2} 45^{\circ}}{2

Vedclass · Accepted Answer

$\sqrt{\frac{3}{2}}$

Vedclass · Answer

$\mathrm{H}_{\max }=\frac{\mathrm{u}^{2} \sin ^{2} 	heta}{2 \mathrm{g}}$

According to problem

$\frac{\mathrm{u}_{1}^{2} \sin ^{2} 45^{\circ}}{2 \mathrm{g}}=\frac{\mathrm{u}_{2}^{2} \sin ^{2} 60^{\circ}}{2 \mathrm{g}}$

$\Rightarrow \frac{\mathrm{u}_{1}^{2}}{\mathrm{u}_{2}^{2}}=\frac{\sin ^{2} 60^{\circ}}{\sin ^{2} 45^{\circ}} \Rightarrow \frac{\mathrm{u}_{1}}{\mathrm{u}_{2}}=\frac{\sqrt{3} / 2}{1 / \sqrt{2}}=\sqrt{\frac{3}{2}}$

Two bodies are thrown up at angles of $45^o$ and $60^o$, respectively, with the horizontal. If both bodies attain same vertical height, then the ratio of velocities with which these are thrown is

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