question_answer

A particle is projected at an angle \[\theta \] from the horizontal with kinetic energy T. What is the kinetic energy of the particle at the highest point?

Answer:

                At highest point, velocity of projectile = \[u\cos \text{ }\!\!\theta\!\!\text{ }\]. K.E. of projectile at highest point = \[\frac{1}{2}m{{\left( u\,\cos \theta  \right)}^{2}}=\left( \frac{1}{2}m{{u}^{2}} \right){{\cos }^{2}}\theta =T{{\cos }^{2}}\theta \]