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

Options

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

Answer

C

Explanation

`E=(1)/(2)mu^(2)`, v = horizontal component of velocity
`= u cos 45^(@) = (u)/(sqrt(2))`
kinetic energy of ball at the highest point
`E'=(1)/(2)mv^(2)=(1)/(2)m((u)/(sqrt(2)))^(2)=(1)/(2)m[(u^(2))/(2)]=(E )/(2)`