HardPhysicsfluid
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-- 43. A person throws a stone in air at some angle. It is observed that after 2 s it is moving at angle 30° to the horizontal and after further 2 s, it is travelling horizontally. The magnitude of initial velocity of stone is
A
20√3 m/s
B
20√7 m/s
C
30√2 m/s
D
15√7 m/s
Explanation
At t=4s, v_y = 0 => u_y - g(4) = 0 => u_y = 40. At t=2s, v_y = u_y - g(2) = 40 - 20 = 20. Also v_y/v_x = tan 30° => 20/v_x = 1/√3 => v_x = 20√3. Initial u = √(u_x² + u_y²) = √((20√3)² + 40²) = √(1200 + 1600) = √2800 = 20√7 m/s.
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