Question

Body is projected at angle 60 degree with horizontal with kinetic energy k when the velocity make an angle 30 degree with horizontal , the kinetic energy of the body will be

Answer 1

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

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

Horizontal component vhorizontal=v∗cos60∘vhorizontal=v∗cos⁡60∘

At the point at which the projectile becomes 30∘30∘

cos30∘=vhorizontalvcos⁡30∘=vhorizontalv

or v=vhorizontalcos30∘v=vhorizontalcos⁡30∘

which gives us v=0.577∗vv=0.577∗v

Kinetic energy at beginning KE=12∗m∗v2KE=12∗m∗v2

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

Therefore kinetic energy when projectile at 30∘30∘becomes 0.3333∗KE0.3333∗KE

Hope it will help you