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At t = 0, an elevator departs from the ground with uniform speed. At time T1 a child drops a marble through the floor. The marble falls with uniform acceleration g = 9.8 m/s^2, and hits the ground T2 seconds later. Find the height of the elevator at time T1. |
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Answer» Let us assume, The velocity with which the elevatar ascends be u', the height attained by elevatar at time,T1 be h . Thus, u' = \(\frac{h}{T_1}\) Equation of motion, h = -u'T2 + \(\frac{1}{2}gT_2^2\) ∵ u' = \(\frac{h}{T_1}\) h + \(\frac{h}{T_1}\)T2 = \(\frac{1}{2}gT_2^2\) h(1 + \(\frac{T_2}{T_1}\)) = \(\frac{1}{2}gT_2^2\) h = \(\frac{1}{2}\)\(\frac{gT_2^2}{(1+\frac{T_2}{T_1})}\) |
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