A particle is projected at ` 60(@)` to the horizontal with a kinetic energy `K` . The kinetic energy at the highest point is

Options

A. zero
B. `(K)/(4)`
C. `(K)/(2)`
D. `K`

Answer

B

Explanation

Let the kinetic energy of the particle at point of projection be `K`.
`K = (1)/(2) m u^(2)`
At highest point, velocity has its horizontal component `u cos theta`.
Therefore kinetic energy of a particle at highet point
Therefore kinetic energy of a particle at highest point is :
`K_(H)=(1)/(2)m(u cos theta)^(2) = (1)/(2) m u^(1) (cos theta)^(2) = K cos^(2) theta`
`rArrK_(H) = K cos^(2) 60^@ = (K)/(4)`.