A ball whose kinetic energy is E, is projected at an angle of 45° to the horizontal. The kinetic energy of the ball at the highest point of its flight will be

- 0
- E
- E/2
- E/√2

**Solution**

Kinetic Energy, E = ½mv^{2}

At the highest point, vertical component of velocity becomes zero and only the horizontal component is left that is given by

u_{x} = u cosθ

θ = 45°, so cosθ = 1/√2

u_{x} = u/√2

Kinetic energy E at the highest point = ½m(u/√2)^{2} = E/2

**The correct option is C.**