Projectile motion in inclined planes?

Question

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 1

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)].