Question

A particle is projected with kinetic energy K at an angle of projection 60°. At highest point the kinetic energy of the particle is:

(a) K

(b) Zero

(c) \(\frac{K}{4}\)

(d) \(\frac{K}{2}\)

Answer 1

Correct option is : (c) \(\frac{K}{4}\)

θ = 60°

\(K = \frac{1}{2}\ mu^2\)

At highest point, v_x = v cos 60° = \(\frac{u}{2}\)

∴ Kinetic energy at highest point,

\(K' = \frac{1}{2}\ mv_x{}^2 = \frac{1}{2}m.\ \frac{u^2}{4} = \frac{1}{4}\times\frac{1}{2}\ mv^2 = \frac{K}{4}\) 

Thus, option (c) is correct.