question_answer

A particle is projected at \[60{}^\circ \]to the horizontal with a kinetic energy K. The kinetic energy at the highest point is

A) K                     

B) zero 

C) \[\frac{K}{4}\]              

D) \[\frac{K}{2}\]

Correct Answer: C

Solution :

[c]: Here, angle of projection, \[\theta =60{}^\circ \] Let u be the velocity of projection of the particle. Kinetic energy of a particle at the point of projection O is\[K=\frac{1}{2}m{{u}^{2}}\]                   ...(i) where m is the mass of a particle. Velocity of the particle at the highest point (i.e. at maximum height) is \[u\cos \theta \]. \[\therefore \]Kinetic energy of the particle at the highest point is \[K'=\frac{1}{2}m{{(ucos\theta )}^{2}}=\frac{1}{2}m{{u}^{2}}{{\cos }^{2}}\theta =\frac{1}{2}m{{u}^{2}}{{\cos }^{2}}{{60}^{o}}\]\[=\frac{1}{2}m{{u}^{2}}{{\left( \frac{1}{2} \right)}^{2}}=\frac{K}{4}\]                  (Using (i))