Two bodies are thrown up at angles of 45^{cir | vedclass

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(C) The maximum vertical height attained by a projectile is given by the formula: $H = \frac{u^{2} \sin^{2} 	heta}{2g}$.Given that both bodies attain

Vedclass · Accepted Answer

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

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(C) The maximum vertical height attained by a projectile is given by the formula: $H = \frac{u^{2} \sin^{2} 	heta}{2g}$.Given that both bodies attain the same vertical height $(H_1 = H_2)$,we have:$\frac{u_{1}^{2} \sin^{2} 45^{\circ}}{2g} = \frac{u_{2}^{2} \sin^{2} 60^{\circ}}{2g}$.Canceling $2g$ from both sides:$u_{1}^{2} \sin^{2} 45^{\circ} = u_{2}^{2} \sin^{2} 60^{\circ}$.Rearranging to find the ratio of velocities $\frac{u_1}{u_2}$:$\frac{u_{1}^{2}}{u_{2}^{2}} = \frac{\sin^{2} 60^{\circ}}{\sin^{2} 45^{\circ}}$.Taking the square root of both sides:$\frac{u_1}{u_2} = \frac{\sin 60^{\circ}}{\sin 45^{\circ}} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{2} 	imes \sqrt{2} = \sqrt{\frac{3}{2}}$.Thus,the ratio of the velocities is $\sqrt{\frac{3}{2}}$.

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

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