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(C) The time of flight for a ball projected vertically upwards with velocity $u_1$ is given by $T_1 = \frac{2u_1}{g}$.The time of flight for a ball pr

Vedclass · Accepted Answer

$\sqrt{3}:2$

Vedclass · Answer

(C) The time of flight for a ball projected vertically upwards with velocity $u_1$ is given by $T_1 = \frac{2u_1}{g}$.The time of flight for a ball projected at an angle $	heta = 60^{\circ}$ with velocity $u_2$ is given by $T_2 = \frac{2u_2 \sin 60^{\circ}}{g}$.Since both balls reach the ground simultaneously,their times of flight are equal,so $T_1 = T_2$.Equating the two expressions: $\frac{2u_1}{g} = \frac{2u_2 \sin 60^{\circ}}{g}$.Simplifying the equation: $u_1 = u_2 \sin 60^{\circ}$.Substituting the value of $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we get $u_1 = u_2 \left( \frac{\sqrt{3}}{2} ight)$.Therefore,the ratio of their velocities is $\frac{u_1}{u_2} = \frac{\sqrt{3}}{2}$.

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