In the figure shown, find.

(a) time of flight of the projectile along the inclined plane.
(b) range OP.

Answer

A, B

Explanation

(a) Horizontal component of initial velocity, `u_x = 20sqrt(2) cos 45^@ = 20 m//s`
Vertical component of initial velocity,
`u_y = 20 sqrt(2) sin 45^@ = 20 m//s`
Let the particle, strikes the inclined plane at P after time t, then horizontal displacement
`QP = u_x t = 20t`
In vertical displacement, `u_yt or 20t ` is upward and `1/2 g t^2 or 5t^2` is downwards. But net vertical
displacement is downwards. Hence `5t^2` should be greater than 20 t and therefore,
`OQ = 5t^2- 20t`
In `DeltaOQP, `
`tan37^@ = OQ/QP`
or `3/4 = (5t^2 - 20t)/(20t)`
Solving this equation, we get
`t= 7s`
(b) Range, OP `= (PQ) sec 37^@`
`= (20t)(5/4)`
Substituting the value of t, we get
`OP = 175 m.`