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

Explanation

Initial kinetic energy ` E = (1)/(2) m u^(2)` . At the highest point , the particle has only the horizontal component of velocity is ucos `45^(@)`
`:. {:("Kinetic"),("energy"):}} E' = (1)/(2) m (u cos 45^(@))^(2)`
`= (1)/(2) mu^(2)((1)/(sqrt(2)))^(2)`
`E' = (E)/(2)`