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Physics · MOTION IN A PLANE
हिन्दीFor a projectile fired from the ground with initial speed \( v_0 \) at an angle \( \theta_0 \) above the horizontal, the ratio of its kinetic energy at the highest point to its initial kinetic energy is \( 1/4 \). What is the time of flight of this projectile in terms of \( v_0 \) and \( g \)?
Correct answer · Option 3
\( \frac{\sqrt{3}v_0}{g} \)
Explanation
\( K_{\text{top}} = K_0 \cos^2\theta_0 = \frac{1}{4} K_0 \implies \cos\theta_0 = 1/2 \implies \theta_0 = 60^\circ \). Time of flight \( T_f = \frac{2 v_0 \sin 60^\circ}{g} = \frac{\sqrt{3} v_0}{g} \).
