A particle is projected making angle 45º with horizontal having kinetic energy K. The kinetic energy at highest point will be

 When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that is directed towards the centre of the earth (we assume that the particle remains close to the surface of the earth). The path of such a particle is called a projectile and the motion is called projectile motion.

Solution

We know that kinetic energy is given by

KE = 1/2 mv2

At heightst point velocity will become u/√2

K’ = 1/2 m +(u/√2)2

K’ = K/2

Answer

The kinetic energy at highest point will be K/2