Physics · MOTION IN A PLANE
हिन्दीThe trajectory equation of a projectile launched from origin is \( y = \sqrt{3}x - \frac{1}{20}x^2 \) where x and y are in meters. Taking \( g = 10\text{ m/s}^2 \), the initial velocity \( v_o \) of projection is:
Correct answer · Option 1
20 m/s
Explanation
Comparing with \( y = x\tan\theta_o - \frac{g}{2v_o^2\cos^2\theta_o}x^2 \): \( \tan\theta_o = \sqrt{3} \implies \theta_o = 60^\circ \). Also \( \frac{g}{2v_o^2\cos^2 60^\circ} = \frac{1}{20} \implies \frac{10}{2v_o^2(1/4)} = \frac{1}{20} \implies v_o^2 = 400 \implies v_o = 20\text{ m/s} \).