# At t = 0, an elevator departs from the ground with uniform speed.

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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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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})}$