Question

A particle is projected at an angle of 60° to the horizontal with a kinetic energy K. The kinetic
energy at the highest point is​

Answer 1

K E = K

=> ½ mv² = K

=> v² = 2K/m

=> v = √(2K/m)

Velocity of projection = √(2K/m)

Horizontal component = √(2K/m) × cos(60)

= ½ √(2K/m)

= √(K/2m)

And the maximum height vertical component of velocity will be zero , and as there is no acceleration in +ve X direction, the horizontal component of velocity will remain the same

So , net velocity at the highest point is √(K/2m)

KE = ½ mv²

= ½ m × K/2m

= K/4