A particle is projected from a point on an inclined plane which makes an angle α above the horizontal. It strikes the inclined plane again with its velocity horizontal. If θ is the angle of projection with the horizontal, then:
Options

Question

A particle is projected from a point on an inclined plane which makes an angle α above the horizontal. It strikes the inclined plane again with its velocity horizontal. If θ is the angle of projection with the horizontal, then:

Options

G1 Physics · Accepted Answer

Horizontal velocity at impact means vᵧ = 0, so the projectile strikes the plane at its highest point. At the highest point, x H x_H x H ​ = u²sinθcosθ/g and H = u²sin²θ/(2g). Since this point lies on the inclined plane, H = x H x_H x H ​ tanα. Therefore [u²sin²θ/(2g)] = [u²sinθcosθ/g]tanα. Cancelling common factors gives tanθ/2 = tanα, hence tanθ = 2tanα. Therefore option B is correct.