Question

A body is projected at an angle of 60 degree with the horizontal ground with kinetic energy k...

Answer 1

Let the initial velocity be v. The projectile starts with 60∘

Now the horizontal component of velocity is same throughout the projectile.

Horizontal component vhorizontal=v∗cos60∘

At the point at which the projectile becomes 30∘

cos30∘=vhorizontalv

or v=vhorizontalcos30∘

which gives us v=0.577∗v

Kinetic energy at beginning KE=12∗m∗v2

Kinetic energy when projectile at 30∘ = KE30∘=12∗m∗(0.577∗v)2

Therefore kinetic energy when projectile at 30∘ becomes 0.3333∗KE