Work, Power and Energy Part 2

Concept:At the highest point, vertical velocity becomes zero, and horizontal velocity remains constant.Kinetic energy at the same horizontal level equals initial kinetic energy due to conservation of mechanical energy.Explanation:Initial kinetic energy at point A: $K_A = K = \frac{1}{2} m u^2$.Angle of projection: $\theta = 60^\circ$.At point B (maximum height), the horizontal velocity is $v_B = u \cos 60^\circ = \frac{u}{2}$.Thus, $K_B = \frac{1}{2} m \left(\frac{u}{2}\right)^2 = \frac{1}{4} \left(\frac{1}{2} m u^2\right) = \frac{K}{4}$.At point C (same level as A), the speed equals the initial speed: $v_C = u$, so $K_C = K$ (conservation of energy).Difference in kinetic energies between B and C: $\Delta K_{BC} = |K_C - K_B| = K - \frac{K}{4} = \frac{3K}{4}$.Required ratio = $\frac{\Delta K_{BC}}{K_A} = \frac{3K/4}{K} = \frac{3}{4}$.Answer:The ratio is $3:4$, which corresponds to option A.