Question

A body is projected at an angle 60 with the horizontal with kinetic energy k what will be kinetic energy whem it make angle 30 with horizontal

Answer 1

when body is projected with speed u at an angle 60° with the horizontal,
then, horizontal component of velocity , $\bf{v_x=ucos60^{\circ}}$
we know, kinetic energy $\propto$v²
so, $\bf{k\propto(vcos60^{\circ})^2}$ ----(1/

Similarly, when a same body is projected at an angle 30° with the horizontal,
then, horizontal component of velocity=$\bf{v'_x=ucos30^{\circ}}$
then, kinetic energy $\propto$(ucos30°)²----(2)

from equations (1) and (2),
kinetic energy/k= u²cos²30°/u²cos²60°
Kinetic wner6 = 3k