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Physics · MOTION IN A STRAIGHT LINE
हिन्दीA ball projected vertically upwards from the edge of a cliff top takes \( t_1 = 2.0 \text{ s} \) to reach its peak height above the cliff. The total time taken by the ball from projection until it strikes the ground at the base of the cliff is \( T = 5.0 \text{ s} \). Taking \( g = 10 \text{ m/s}^2 \), calculate the height of the cliff.
Correct answer · Option 3
\( 25 \text{ m} \)
Explanation
Time to maximum height \( t_1 = \frac{v_0}{g} = 2.0 \implies v_0 = 20 \text{ m/s} \). Total vertical displacement from cliff top to ground: \( y = v_0 T - \frac{1}{2} g T^2 = 20(5) - 5(25) = 100 - 125 = -25 \text{ m} \). Thus, the height of the cliff is \( 25 \text{ m} \).
