question_answer

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

Answer:

                  If the particle is projected with velocity \[\upsilon \], then its initial kinetic energy is \[K=\frac{1}{2}m{{\upsilon }^{2}}\] Velocity of the projectile at the highest point \[=\upsilon \cos \theta \] \[\therefore \]K.E. of the particle at the highest point \[=\frac{1}{2}m{{(\upsilon \cos \theta )}^{2}}=(\frac{1}{2}m{{\upsilon }^{2}}){{\cos }^{2}}\theta =K{{\cos }^{2}}\theta .\]