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Two bodies of masses m1 and m2 are let fall freely from heights 1 h and 2 h respectively. The ratio of time taken by the bodies to fall through these heights is |
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Answer» nTwo bodies of masses m1 and m2 are fallen freely from heights h1 and h2 respectively. The ratio of time taken by the bodies to fall through these heights is.SolutionFor first body:MASS (m) = m1Height (H) = h1For SECOND body:Mass (m) = m2Height (h) = h2Given that, Two bodies of masses m1and m2 are fallen freely from heights h1 and h2 respectively.For the freely FALLING body, a = g = 9.8 m/s² or 10 m/s² (approx.)Initial velocity (u) = 0 m/sUsing Second Equation of Motion i.e.s= ut + 1/2 at²Assume that, the first body takes t1 time and second body takes t2 time.For the first body:h1 = 0(t1) + 1/2 g(t1)²h1 = 1/2 g(t1)²2(h1) = g(t1)²[2(h1)]/g = (t1)²t1 = √[2(h1)/g]For the second body:h2 = 1/2 g(t2)²t2 = √[2(h2)/g]We have to find the ratio of time taken by the bodies to fall through these heights.→ t1/t2 = [√2h1/g] / [√2h2/g]→ t1/t2 = √h1/√h2 |
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