Projectile motion on inclined plane

Consider a plane inclined at an angle θₒ with horizontal. Let a projectile be projected from the foot of the inclined plane A, with velocity v and at an angle as shown in figure. The particle strikes the plane at point B. We want to find the range of projectile on the inclined plane which is R = AB.

Take x – axis parallel to the inclined plane and y - axis.

Perpendicular to the Inclined plane as shown we can write the following.

u_x = u cos α, u_y = u sin α,

a_x = - g sin θₒ   a_y = - g cos θₒ

From A to B S_x = R, S_y = 0

And let time taken from A to B be T.

From A to B applying

S_y = u_yt + ½ ayt²

We get

0 = (u sin α) T - ½ g cos θₒT²

T = 2u sinα/g cos θₒ… (i)

Now applying

S_x = u_xt + ½ axt²

R = (u cos α) T - ½ g sin θₒT² ... (ii)

Putting the value of T from (i) in (ii) we get

R = u cos α (2u sin α/ g cosθₒ) T - ½ g sin θₒ (2u sin α/ g cosθₒ)²

R = (2u² cos α sin α/ g cosθₒ) - (2u²sin²α g sin θₒ/ g cos²θₒ)

R = u²/ g cos²θₒ (sin²α cos θₒ - sin θₒ 2 sin²α)

R = u²/ g cos²θₒ [sin (2α + θₒ) - sin θₒ].