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(A) For collision,the vertical components of the velocities of the ball and the stone must be equal,so that they remain at the same height throughout

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) For collision,the vertical components of the velocities of the ball and the stone must be equal,so that they remain at the same height throughout the flight.Case $I$: $(	heta_0, 	heta_1) = (45^{\circ}, 45^{\circ})$.Vertical components: $v_0 \sin 45^{\circ} = v \sin 45^{\circ} \implies v = v_0$.The horizontal components of velocity are $v_{0x} = v_0 \cos 45^{\circ} = \frac{v_0}{\sqrt{2}}$ and $v_{sx} = v \cos 45^{\circ} = \frac{v}{\sqrt{2}} = \frac{v_0}{\sqrt{2}}$.The relative horizontal speed is $v_{rel} = v_{0x} + v_{sx} = \frac{v_0}{\sqrt{2}} + \frac{v_0}{\sqrt{2}} = \sqrt{2}v_0$.Time $T_1 = \frac{L}{v_{rel}} = \frac{L}{\sqrt{2}v_0}$.Case $II$: $(	heta_0, 	heta_1) = (60^{\circ}, 30^{\circ})$.Vertical components: $v_0 \sin 60^{\circ} = v \sin 30^{\circ} \implies v_0 \frac{\sqrt{3}}{2} = v \frac{1}{2} \implies v = \sqrt{3}v_0$.The horizontal components of velocity are $v_{0x} = v_0 \cos 60^{\circ} = \frac{v_0}{2}$ and $v_{sx} = v \cos 30^{\circ} = (\sqrt{3}v_0) \frac{\sqrt{3}}{2} = \frac{3v_0}{2}$.The relative horizontal speed is $v_{rel} = v_{0x} + v_{sx} = \frac{v_0}{2} + \frac{3v_0}{2} = 2v_0$.Time $T_2 = \frac{L}{v_{rel}} = \frac{L}{2v_0}$.Now,$(T_1 / T_2)^2 = \left( \frac{L / (\sqrt{2}v_0)}{L / (2v_0)} ight)^2 = \left( \frac{2v_0}{\sqrt{2}v_0} ight)^2 = (\sqrt{2})^2 = 2$.

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