A stone is thrown at an angle of `45^(@)` to the horizontal with kinetic energy K. The kinetic energy at the highest point is

Options

A. `(K)/(sqrt2)`
B. `(K)/(2)`
C. `2K`
D. `k`

Answer

B

Explanation

(b): A particle is projected at an angle `45^(@)` with horizontal . The velocity at highest point is
`u_(y)=0, u_(x) u cos 45^(@)`
`u_(x)=(u)/(sqrt2)`
Kinetic energy of particle having velocity u
`K=(1)/(2)m u^(2)`
Now kinetic energy at highest point is K
`K=(1)/(2)m v_(x)^(2)=(1)/(2) mu^(2)((1)/(2))`
`K=(K)/(2)`