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 = 20(sqrt2) cos 45^@ = 20 m//s`
Vertical component of initial velocity
`u_y = 20(sqrt2) sin 45^@ = 20 m//s`
Let the particle strikes at P after time t, then
horizontal displacement `OQ = u_xt = 20t `
In vertical displacement, `u_yt` or 20t is upwards and `1/2 g t^2` or `5t^2` is downwards. But net
displacement is upwards, therefore 20 t should be greater than `5t^2` and
`QP = 20t - 5t^2 `
In `Delta OPQ, `
`tan 37^@ = (QP/OQ)`
or, ` 3/4 = ((20t - 5t^2)/(20t))`
Solving this equation, we get
`t = 1s`
(b) Range `OP = OQ sec 37^@`
` =(20t)(5/4)`
Substituting t = 1s, we have
OP = 25m .