Question

A particle is projected in air at an angle β to a surface which itself is inclined at an angle α to the horizontal (Fig.)

(a) Find an expression of range on the plane surface (distance on the plane from the point of projection at which particle will hit the surface).

(b) Time of flight.

(c) β at which range will be maximum.

[Hint : This problem can be solved in two different ways:

(i) Point P at which particle hits the plane can be seen as intersection of its trajectory (parabola) and straight line. Remember particle is projected at an angle (α + β) w.r.t. horizontal.

(ii) We can take x-direction along the plane and y-direction perpendicular to the plane. In that case resolve g (acceleration due to gravity) two different components, along the plane and perpendicular to the plane. Now the problem can be solved as two independent motions in x and y direction respectively with time as a common parameter.)

Answer 1

Part (b) solved first :

(b) From A to B,

y = 0, u_y= v₀ sin β,

a_y= −g cos α, t = T

y =u_yt + \(\frac{1}{2}\)a_yt²

0 = v₀sin β.T + \(\frac{1}{2}\)(− cos α) T²

⇒T = \(\frac{2v_0sin\beta}{gcos\alpha}\)

(a) x = L, u= cos β

a_x= −g sin α

x =u_xt + \(\frac{1}{2}\)a_xt ²

L = v₀cos βT + \(\frac{1}{2}\) (− sin α) T²

L = T[v₀ cos β − \(\frac{1}{2}\) g sin α × \(\frac{2v_0sin\beta}{gcos\alpha}\)]

= \(\frac{2v^2_0sin\beta}{g sin^2\alpha}\) [cos β cos α − sin α. sin β ]

⇒ L = \(\frac{2v^2_0 sin\beta}{g sin^2\alpha}\)cos (α + β)