A particle is projected at $${60^ \circ }$$  to the horizontal with a kinetic energy $$K.$$  The kinetic energy at the highest point is-

Solution :
Let $$u$$  be the velocity with which the particle is thrown and $$m$$  be the mass of the particle. Then
$$K = \frac{1}{2}m{u^2}\,.....(1)$$
At the highest point the velocity is $$u\,cos\,{60^ \circ }$$   (only the horizontal component remains, the vertical component being zero at the top-most point). Therefore kinetic energy at the highest point.
$$\eqalign{ & K' = \frac{1}{2}m{\left( {u\,cos\,{{60}^ \circ }} \right)^2} \cr & = \frac{1}{2}m{u^2}co{s^2}{60^ \circ } \cr & = \frac{K}{4}\,\left[ {{\text{From }}1} \right] \cr} $$