Question

the angle for which kinetic energy at highest point is equal to one fourth of its kinetic energy at point of projection?

Answer 1

Let £ be the angle of projection with ground.
Therefore at the top, velocity=ucos£
where u is initial angle of projection

Now,

At the start of the motion
u = ux i + uy j. where i and j are unit vectors along x-axis and y-axis

Therefore u=ucos£ i + usin£ j
ATQ

K.E at highest point=1/2m(ucos£) ^2
k.e at lowest point=1/2m(ucos£^2 + usin£^2)

Now K. E=1/4 ke
1/2m(ucos£)^2 =1/8m(ucos£^)2+(usin£^2)
(ucos£) ^2 =1/4(ucos£)^2 +(usin£) ^2

3(ucos£)^2 =(usin£) ^2
(tan£)^2 =3
therefore tan£=
$\sqrt{3 }$
Hence £=60°