Question

A projectile is thrown with a Speed u,at an angle theta to an inclined plane of inclination beeta.the angel theta at which the projectile is thrown such that it strikes the inclined plane normally

Answer 1

Answer:

From resolved vector components, we have,

v/u cos θ = tan(90- β) = cotβ or, v^2= u^2cos^ θcot^2β...............(1)

where v is the vertical component of velocity vector at point A, u is the projection velocity. Refer the figure for angles.

From projectile motion,

v^2= u^2 sin^θ - 2g (Lsin β)....................(2)

from (1) and (2), we can write

u^2cos^ θcot^2β= u^2 sin^θ - 2g (Lsin β).......................(3)

Eqn.(3) is the relation between u, L, β and θ . Using trignometry and algebrical simplification

θ= sin^(-1)√ ( cos^β + 2gL/u^2 sin^2β)