Question

A particle is projected at an angle theta from the horizontal with kinetic energy k

Answer 1

Given that angle formed by the projectile and the ground is θθ and Kinetic energy KK

We know a simple fact that,

Kinetic Energy, K=½mv2K=½mv2 ___________(1)

At the highest point, vertical component of velocity becomes zero and only the horizontal component is left that is given by :

v(x)=vcosθv(x)=vcosθ ___________(2)

Substituting (2) in (1),

→ KE at the highest point = ½m(vcosθ)2½m(vcosθ)2

→ KE at the highest point = ½mv2(cosθ)2½mv2(cosθ)2

→Kinetic energy at the highest point = K(cosθ)2K(cosθ)2 (From (1),as, K=½mv2K=½mv2)

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