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(D) The formula for the maximum height $H$ reached by a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$, where $u$ is the initial velocity a

Vedclass · Accepted Answer

$\sqrt{3}:\sqrt{2}$

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(D) The formula for the maximum height $H$ reached by a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$, where $u$ is the initial velocity and $	heta$ is the angle of projection.Given that the maximum heights reached by both balls $A$ and $B$ are the same, we have $H_A = H_B$.Substituting the formula, we get $\frac{u_A^2 \sin^2(45^{\circ})}{2g} = \frac{u_B^2 \sin^2(60^{\circ})}{2g}$.Canceling $2g$ from both sides, we get $u_A^2 \sin^2(45^{\circ}) = u_B^2 \sin^2(60^{\circ})$.Substituting the values $\sin(45^{\circ}) = 1/\sqrt{2}$ and $\sin(60^{\circ}) = \sqrt{3}/2$, we get $u_A^2 (1/\sqrt{2})^2 = u_B^2 (\sqrt{3}/2)^2$.This simplifies to $u_A^2 (1/2) = u_B^2 (3/4)$.Rearranging the terms, $\frac{u_A^2}{u_B^2} = \frac{3/4}{1/2} = \frac{3}{4} 	imes 2 = \frac{3}{2}$.Taking the square root of both sides, the ratio of the initial velocities is $\frac{u_A}{u_B} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}}$.

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