A particle is projected from ground from point A near the bottom of a wedge. At the same time wedge (sufficiently long) is also moved with same speed towards right as shown, then

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Particle can never collide the wedge.

Particle will collide the wedge after 2 sec.

Maximum height attained by the particle is 11.25 m from the ground.

Height of point on the wedge where particle strikes is 10m.

answer is B, C, D.

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Detailed Solution

$\frac{y}{x} = \tan 30 ° = \frac{(10 \sqrt{3}) \sin 60^{0} t - \frac{1}{2} g t^{2}}{V_{w} t - (10 \sqrt{3} \cos 60^{0}) t} \Rightarrow \frac{1}{\sqrt{3}} = \frac{15 - 5 t}{10 \sqrt{3} - 5 \sqrt{3}} S o l v i n g t = 2 s y_{m a x} = \frac{{(10 \sqrt{3})}^{2} \sin^{2} 60^{0}}{2 g} = \frac{45}{4} = 11.25 m y_{s t r i k e} = 10 \sqrt{3} \sin 60^{0} t - \frac{1}{2} g t^{2} = (10 \sqrt{3}) (\frac{\sqrt{3}}{2}) (2) - \frac{1}{2} (10) (4) = 10 m$