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A man release a stone at the top edge of a tower. During the last second of its travel,the stone fall through a distance of(9/25)H,where H is the tower’s height .Find H
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|dw:1443817879110:dw| equations of motion (EoM) use EoM \(v^2 = u^2 + 2ax\) to establish velocities at positions shown in blue in drawing from EoM \(v = u + at\) for final 1 second of flight time, we can say \(v = u + g(1)\) \(\sqrt{2gH} = \sqrt{2g \frac{16}{25}H} + g\) then, solve for H [in terms of g]
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