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Physics · MOTION IN A PLANE
हिन्दीA body is projected from ground level with speed \( v_0 \) at an angle \( \theta_0 \) to the horizontal. If the horizontal range \( R \) is four times the maximum height \( H_m \) achieved by the body (\( R = 4 H_m \)), what is the angle of projection \( \theta_0 \)?
Correct answer · Option A
\( 45^\circ \)
Explanation
\( R = \frac{2v_0^2\sin\theta_0\cos\theta_0}{g} \) and \( H_m = \frac{v_0^2\sin^2\theta_0}{2g} \). \( R = 4H_m \implies \frac{2v_0^2\sin\theta_0\cos\theta_0}{g} = \frac{2v_0^2\sin^2\theta_0}{g} \implies \tan\theta_0 = 1 \implies \theta_0 = 45^\circ \).
