A particle is projected at an angle of `60^(@)` to the horizontal with a kinetic energy E. The kinetic energy at the highest point is

Answer

A, B, C

Explanation

Let v the velocity at the point where it makes an angle `theta/` with horizontal.
The horizontal component remains unchanged.
So, `v cos (theta/2)=ucostheta`

`rarr v=(u costheta)/(cos(theta/2))`…….i
From ure, `mgcos(theta/2)=(mv^2)/r`
`rarr r=v^2/(gcos(theta/2))`
putting the value of v from equation i.
`r=(v^2cos^2theta)/(g cos^2(theta/2))`