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16.A balloon rises from rest on the ground with constant acceleration 2. A stone is dropped when theballoon has rises to a height 60 metre. The time taken by the stone to reach the ground is.Motion with variable acceleration & calculus |
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Answer» Answer: Explanation: Answer: Explanation: Velocity after rising height h v = sqrt(2aS) ………[∵ initial velocity of balloon is zero] v = sqrt(2 × G/8 × h) v = sqrt(gh/4) This is the initial velocity of stone. Time taken by stone rise extra height and come to it’s point of PROJECTION is t1 = 2U / g = 2 × sqrt(gh/4) / g = sqrt(h / g) Final velocity of stone when it TOUCHES the ground v’ = sqrt(u^2 + 2aS) = sqrt[(gh/4) + 2gh] = sqrt[2.25 gh] = 1.5 × sqrt(gh) Time taken by stone to FALL down from point of projection t2 = (v’ - u) / a = [(1.5 × sqrt(gh)) - sqrt(gh/4)] / g = sqrt(gh) [1.5 - 0.5] / g = sqrt(h / g) Total time taken for stone to reach the ground T = t1 + t2 = 2 sqrt(h / g) |
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