Question

a particle of mass m is projected with velocity v making an angle 45 degree with the horizontal having kinetic energy k the kinetic energy at highest point will be

Answer 1

Heya!

Given,

Φ = 45°

velocity of the particle in the horizontal direction (v) = vcos Φ

$\boxed{\pink{Kinetic \: Energy = \dfrac{1}{2} m{v}^{2}}}$

K.E. = $\dfrac{1}{2}m{v}^{2}$

K.E. = $\dfrac{1}{2}m{(v cos 45)}^{2}$

K.E. = $\dfrac{1}{2}m{v}^{2}\dfrac{1}{2}$

K.E. = $\dfrac{1}{4}m{v}^{2}$