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Physics · MOTION IN A PLANE
हिन्दीA hiker stands on the edge of a cliff \( 490\text{ m} \) above the ground and throws a stone horizontally with an initial speed of \( 15\text{ m/s} \). Taking \( g = 9.8\text{ m/s}^2 \), the time taken by the stone to reach the ground and the speed with which it hits the ground are respectively:
Correct answer · Option 4
\( 10\text{ s} \) and \( 99\text{ m/s} \)
Explanation
\( y = \frac{1}{2} g t^2 \implies 490 = 4.9 t^2 \implies t = 10\text{ s} \). \( v_x = 15\text{ m/s} \), \( v_y = gt = 98\text{ m/s} \). Total speed \( v = \sqrt{15^2 + 98^2} = \sqrt{225 + 9604} = \sqrt{9829} \approx 99\text{ m/s} \).
