question_answer

At which points on the projectile trajectory is the (i) potential energy maximum (ii) kinetic energy minimum and (iii) total energy maximum?

Answer:

                (i) P.E. of a projectile is maximum at its highest point because of its maximum height. It is given by \[{{(P.E.)}_{H}}=mg\,H=mg.\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\frac{1}{2}m{{u}^{2}}{{\sin }^{2}}\theta \]  (ii) K.E. of a projectile is minimum (not zero) at its highest point because of its minimum velocity. \[{{(K.E.)}_{H}}=\frac{1}{2}mu_{H}^{2}=\frac{1}{2}m{{(u\cos \theta )}^{2}}\]                 \[=\frac{1}{2}m{{u}^{2}}{{\cos }^{2}}\theta \]  (ii) Total energy at highest point \[={{(P.E.)}_{H}}+{{(K.E)}_{H}}\] \[=\frac{1}{2}m{{u}^{2}}({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )=\frac{1}{2}m{{u}^{2}}\] \[=\] Total energy at the point of projection Hence total energy of a projectile is conserved at all points of its motion. The K.E. is maximum at the point of projection.