Two inclined planes I and II have inclination $α$ and $β$ respectively with horizontal, (where $α + β = 90^{0}$ ) intersect each at point ‘O’ as shown in figure. A particle is projected from point A with velocity ‘u’ along a direction perpendicular to plane (I). If the particle strikes plane (II) perpendicularly at B, then

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Time of flight = $\frac{u}{g \sin β}$ .

Distance $O B = \frac{u^{2}}{2 g \sin α}$

Time of flight = $\frac{u}{g \sin α}$

Distance $O B = \frac{u^{2}}{2 g \sin β}$

answer is A, C.

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Detailed Solution

$u = g \sin β t$

$t = \frac{u}{g \sin β} = time of flight$

$O B = u \frac{u}{g \sin β} - \frac{1}{2} g \sin β \frac{u^{2}}{g^{2} \sin^{2} β}$

= $\frac{u^{2}}{g \sin β} - \frac{u^{2}}{2 g \sin β}$

$O B = \frac{u^{2}}{2 g \sin β}$

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