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Physics · MOTION IN A PLANE
हिन्दीA particle thrown vertically upwards from ground reaches a maximum height \( H \). The ratio of the time taken by the particle to reach height \( \frac{3}{4}H \) during its upward motion to the total time of flight \( T_f \) is:
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Correct answer · Option B
\( 1/4 \)
Explanation
\( y = u_y t - \frac{1}{2}gt^2 = \frac{3}{4}(\frac{u_y^2}{2g}) \implies 4(\frac{gt}{u_y})^2 - 8(\frac{gt}{u_y}) + 3 = 0 \implies \frac{gt}{u_y} = \frac{1}{2} \). Thus \( t = \frac{u_y}{2g} \). Since \( T_f = \frac{2u_y}{g} \), the ratio is \( \frac{t}{T_f} = \frac{1}{4} \).
