JEE Main & Advanced Physics Two Dimensional Motion Projectile Motion on An Inclined Plane

Let a particle be projected up with a speed u from an inclined plane which makes an angle \[\alpha \] with the horizontal and velocity of projection makes an angle \[\theta \] with the inclined plane.

We have taken reference x-axis in the direction of plane.

Hence the component of initial velocity parallel and perpendicular to the plane are equal to \[u\cos \theta \] and \[u\sin \theta \] respectively i.e. \[{{u}_{||}}=u\cos \theta \] and  \[{{u}_{\bot }}=u\sin \theta \].

The component of g along the plane is \[g\sin \alpha \] and perpendicular to the plane is \[g\cos \alpha \] as shown in the figure i.e. \[{{a}_{||}}=-g\sin \alpha \] and \[{{a}_{\bot }}=g\cos \alpha \].  

Therefore the particle decelerates at a rate of \[g\sin \alpha \] as it moves from O to P.  

(1) Time of flight : We know for oblique projectile motion \[T=\frac{2u\sin \theta }{g}\]or    we can say  \[T=\frac{2{{u}_{\bot }}}{{{a}_{\bot }}}\]

\[\therefore \] Time of flight on an inclined plane \[T=\frac{2u\sin \theta }{g\cos \alpha }\]

(2) Maximum height : We know for oblique projectile motion \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]      or  we can say \[H=\frac{u_{\bot }^{2}}{2{{a}_{\bot }}}\]

\[\therefore \] Maximum height on an inclined plane \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g\cos \alpha }\]

(3) Horizontal range : For one dimensional motion \[s=ut+\frac{1}{2}a{{t}^{2}}\]

Horizontal range on an inclined plane \[R={{u}_{||}}T+\frac{1}{2}{{a}_{||}}{{T}^{2}}\] \[R=u\cos \theta \,T-\frac{1}{2}g\sin \alpha \,\,{{T}^{2}}\]           

\[R=u\cos \theta \,\left( \frac{2u\sin \theta }{g\cos \alpha } \right)-\frac{1}{2}g\sin \alpha \,{{\left( \frac{2u\sin \theta }{g\cos \alpha } \right)}^{2}}\]

By solving   \[R=\frac{2{{u}^{2}}}{g}\,\frac{\sin \theta \,\cos (\theta +\alpha )}{{{\cos }^{2}}\alpha }\]

(i) Maximum range occurs when \[\theta =\frac{\pi }{4}-\frac{\alpha }{2}\]

(ii) The maximum range along the inclined plane when the projectile is thrown upwards is given by           

\[{{R}_{\max }}=\frac{{{u}^{2}}}{g\,(1+\sin \alpha )}\]

(iii) The maximum range along the inclined plane when the projectile is thrown downwards is given by

\[{{R}_{\max }}=\frac{{{u}^{2}}}{g\,(1-\sin \alpha )}\]