A cricket ball is hit at an angle 45° to the horizontal with kinetic energy K the kinetic energy at the highest point is what?

a)k

b)k/2

c)ksin 45°

d)zero

Kinetic Energy, K= ½mv^2

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

ux = u cosθ

θ = 45°, so cosθ = 1/√2

ux = u/√2

Kinetic energy K at the highest point = ½m(u/√2)^2 = (1/2 ×(mu^2))/2
=K/2