Question

A cricket ball of mass m is hitted by an angle 45° to the horizontal with velocity v its kinetic energy at topmost point is

Answer 1

Answer:

K.E' = K.E/2

Explanation:

K.E = ½ mv²

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 cos60° = 1/√2

ux = u/2

K.E at the highest point =  ½ m (u/√2)² = K.E/2