NEET Physics Two Dimensional Motion Projectile Motion on an Inclined Plane

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 velocity of projection makes an angle q 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 \[T=\frac{2{{u}_{\bot }}}{{{a}_{\bot }}}\] \[\therefore \,\,\,\,Time\text{ }of\text{ }flight\text{ }on\text{ }an\text{ }inclined\text{ }plane\,\,T=\frac{2u\sin \theta }{g\cos \alpha }\] (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 \[{{R}_{\max }}=\frac{{{u}^{2}}}{g\,(1-\sin \alpha )}\] Sample problem based on inclined projectile Problem 77. For a given velocity of projection from a point on the inclined plane, the maximum range down the plane is three times the maximum range up the incline. Then, the angle of inclination of the inclined plane is (a) \[{{30}^{o}}\]        (b) \[{{45}^{o}}\]        (c) \[{{60}^{o}}\]                     (d) \[{{90}^{o}}\] Solution: (a) Maximum range up the inclined plane \[{{({{R}_{\max }})}_{up}}=\frac{{{u}^{2}}}{g(1+\sin \alpha )}\] Maximum range down the inclined plane \[{{({{R}_{\max }})}_{down}}=\frac{{{u}^{2}}}{g(1-\sin \alpha )}\] and according to problem: \[\frac{{{u}^{2}}}{g(1-\sin \alpha )}=3\times \frac{{{u}^{2}}}{g(1+\sin \alpha )}\] By solving \[\alpha =\text{ }{{30}^{o}}\] Problem 78. A shell is fired from a gun from the bottom of a hill along its slope. The slope of the hill is \[\alpha =\text{ }{{30}^{o}}\], and the angle of the barrel to the horizontal \[\beta = 6{{0}^{o}}\]. The initial velocity v of the shell is 21 m/sec. Then distance of point from the gun at which shell will fall (a) 10 m            (b) 20 m            (c) 30 m            (d) 40 m Solution: (c) Here u = 21 m/sec, \[\alpha = 3{{0}^{o}},\,\,q=\beta \,\alpha = 6{{0}^{o}} 3{{0}^{o}}= 3{{0}^{o}}\] Maximum range \[R=\frac{2{{u}^{2}}}{g}\,\frac{\sin \theta \,\cos (\theta +\alpha )}{{{\cos }^{2}}\alpha }=\frac{2\,\times {{(21)}^{2}}\times \sin {{30}^{o}}\cos {{60}^{o}}}{9.8\times {{\cos }^{2}}{{30}^{o}}}=30\,\,m\] Problem 79. The maximum range of rifle bullet on the horizontal ground is 6 km its maximum range on an inclined of 30o will be (a) 1 km            (b) 2 km (c) 4 km             (d) 6 km Solution: (c) Maximum range on horizontal plane \[R=\frac{{{u}^{2}}}{g}=6\,km\] (given) Maximum range on a inclined plane \[{{R}_{\max }}=\frac{{{u}^{2}}}{g\,(1+\sin \alpha )}\] Putting \[\alpha = 3{{0}^{o}}\]   \[{{R}_{\max }}=\frac{{{u}^{2}}}{g\,(1+\sin {{30}^{o}})}=\frac{2}{3}\left( \frac{{{u}^{2}}}{g} \right)=\frac{2}{3}\times 6=4\,km.\]