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Physics · MOTION IN A PLANE
हिन्दीA body projected at an angle \( \theta_0 \) to the horizontal follows a parabolic trajectory given by \( y = \sqrt{3}x - \frac{g x^2}{20} \) (where \( x, y \) are in metres and \( g = 10\text{ m/s}^2 \)). The angle of projection \( \theta_0 \) and the initial speed \( v_0 \) are respectively:
Correct answer · Option 1
\( 60^\circ \) and \( 20\text{ m/s} \)
Explanation
Comparing with \( y = (\tan\theta_0) x - \frac{g x^2}{2 v_0^2 \cos^2\theta_0} \): \( \tan\theta_0 = \sqrt{3} \implies \theta_0 = 60^\circ \). Also \( \frac{g}{2 v_0^2 \cos^2 60^\circ} = \frac{g}{20} \implies \frac{10}{2 v_0^2 (1/4)} = \frac{10}{20} \implies v_0^2 = 400 \implies v_0 = 20\text{ m/s} \).
