A cricket ball is hit for a six the bat at an angle of `45^(@)` to the horizontal with kinetic energy K. At the highest point, the kinetic energy of the ball is

Options

A. zero
B. k
C. k/2
D. `k//sqrt(2)`

Answer

c

Explanation

`(1)/(2)mu^(2)=k`
At top velocity is `((u)/sqrt(2))`
Hence K.E. `=(1)/(2)(mu^(2))/(2)=((K)/(2))`