# Projectile Motion on Inclined Plane

Consider a plane inclined at an angle θ0 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.

ux = u cos α, uy = u sin α,

ax = – g sin θₒ   ay = – g cos θₒ

From A to B Sx = R, Sy = 0

And let time taken from A to B be T.

From A to B applying

Sy = uyt + ½ ayt²

We get

$$0=(u\,\sin \alpha )T-\,\frac{1}{2}g\text{ co}s{{\theta }_{0}}T{}^\text{2}$$

T = 2u sinα/g cos θₒ… (i)

Now applying

Sx = uxt + ½ axt²

$$R=(u\text{ }cos\alpha )T-\,\frac{1}{2}g\text{ }sin{{\theta }_{0}}T{}^\text{2}$$ … (ii)

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

$$R=u\text{ }cos\alpha \left( \frac{2u\text{ }sin\text{ }\alpha }{g\text{ }cos{{\theta }_{0}}} \right)T-\frac{1}{2}g\text{ }sin{{\theta }_{0}}{{\left( \frac{2u\text{ }sin\text{ }\alpha }{g\text{ }cos{{\theta }_{0}}} \right)}^{2}}$$

$$R=\left( \frac{2u{}^\text{2}cos\alpha \text{ }sin\text{ }\alpha }{g\text{ }cos{{\theta }_{0}}} \right)-\left( \frac{2u{}^\text{2}sin{}^\text{2}\alpha g\text{ }sin{{\theta }_{0}}}{g\text{ }co{{s}^{2}}{{\theta }_{0}}} \right)$$

$$R\text{ }=\frac{u{}^\text{2}}{g\text{ }cos{}^\text{2}{{\theta }_{0}}}\left( sin{}^\text{2}\alpha \text{ }cos\text{ }{{\theta }_{0}}-\text{ }sin{{\theta }_{0}}2\text{ }sin{}^\text{2}\alpha \right)$$

$$R\text{ }=\frac{u{}^\text{2}}{g\text{ }cos{}^\text{2}{{\theta }_{0}}}\left( sin\left( 2\alpha +{{\theta }_{0}} \right)-sin{{\theta }_{0}} \right)$$