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Physics · MOTION IN A PLANE
हिन्दीA particle is projected from ground level with speed \( u \) at an angle \( \theta \) with the horizontal. At what fraction of its maximum height \( h_m \) is the vertical component of its velocity equal to half of its initial vertical component?
Correct answer · Option 3
\( 3/4 \)
Explanation
\( v_y^2 = v_{0y}^2 - 2gh \implies (\frac{u\sin\theta}{2})^2 = (u\sin\theta)^2 - 2gh \implies 2gh = \frac{3}{4} u^2 \sin^2\theta \). Since \( h_m = \frac{u^2\sin^2\theta}{2g} \), we get \( h = \frac{3}{4} h_m \).
