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