question_answer

A ball whose kinetic energy is E, is projected at an angle of \[45{}^\circ \] to the horizontal. The kinetic energy of the ball at the highest point of its flight will be

A) E

B) \[E/\sqrt{2}\]

C) \[E/2\]

D) zero

Correct Answer: C

Solution :

[c] Let u be the speed with which the ball of mass m is projected. Then the kinetic energy (E) at the point of projection is \[E=\frac{1}{2}m{{u}^{2}}\]   ?(i) When the ball is at the highest point of its flight, the speed of the ball is \[\frac{u}{\sqrt{2}}\] (Remember that the horizontal component of velocity does not change during a projectile motion). \[\therefore \]The kinetic energy at the highest point \[=\frac{1}{2}m{{\left( \frac{u}{\sqrt{2}} \right)}^{2}}=\frac{1}{2}\frac{m{{u}^{2}}}{2}=\frac{E}{2}\][From (i)]