Q.279. Projectile motion from and to a pair of intersecting inclined planes.

Question 279.
Two inclined planes intersect in a horizontal plane. Their inclination to the horizontal is α and β. If a particle is projected at right angles to the former from a point in it so as to strike the other at right angles, then find the velocity of projection. Assume that the particle undergoes a vertical displacement of 'h' during its motion.

Answer 279.
Let the first inclined plane be at an obtuse angle 'α' and the other plane be at an acute angle 'β'.
Let v be the velocity of projection and u be the velocity of landing.

As the horizontal component of both velocities must be the same
v cos(α - 90°) = u cos(90° - β)
=> v sinα = u sinβ
=> u² sin² β - v² sin² α = 0 ... (1)

For vertical motion,
[usin(90° - β)]² - [vsin(α - 90°)]² = 2gh
=> u² cos² β - v² cos² α = 2gh ... (2)

Adding eqns. (1) and (2),
u² - v² = 2gh
=> u² = v² + 2gh

Plugging this value of u² in eqn. (1),
(v² + 2gh) sin² β - v² sin² α = 0
=> v² = 2gh sin² β / (sin² α - sin² β)
=> v = sin β * √[2gh/(sin² α - sin² β)].

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