Two objects are thrown up at angles of 45^{ci | vedclass

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

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

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(D) The maximum vertical height $H$ attained by a projectile is given by the formula $H = \frac{u^2 \sin^2 	heta}{2g}$.Given that both objects attain the same vertical height,we have $H_1 = H_2$.Substituting the formula,we get $\frac{u_1^2 \sin^2 	heta_1}{2g} = \frac{u_2^2 \sin^2 	heta_2}{2g}$.Canceling $2g$ from both sides,we get $u_1^2 \sin^2 	heta_1 = u_2^2 \sin^2 	heta_2$.Given $	heta_1 = 45^{\circ}$ and $	heta_2 = 60^{\circ}$,we have $u_1^2 \sin^2 45^{\circ} = u_2^2 \sin^2 60^{\circ}$.Substituting the values $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$ and $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we get $u_1^2 \left(\frac{1}{\sqrt{2}}ight)^2 = u_2^2 \left(\frac{\sqrt{3}}{2}ight)^2$.This simplifies to $u_1^2 \left(\frac{1}{2}ight) = u_2^2 \left(\frac{3}{4}ight)$.Rearranging for the ratio $\frac{u_1^2}{u_2^2}$,we get $\frac{u_1^2}{u_2^2} = \frac{3/4}{1/2} = \frac{3}{4} 	imes 2 = \frac{3}{2}$.Taking the square root on both sides,the ratio of the velocities is $\frac{u_1}{u_2} = \sqrt{\frac{3}{2}}$.

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

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