Question
An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 6ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.

Answer

For elevator and coin u = 0 As the elevator descends downward with acceleration a' (say) The coin has to move more distance than 1.8m to strike the floor. Time taken t = 1 sec.$\text{S}_{\text{c}}=\text{ut}+\frac{1}{2}\text{a}'\text{t}^2=0+\frac{1}{2}\text{g}(1)^2=\frac{1}{2}\text{g}$
$\text{S}_{\text{e}}=\text{ut}+\frac{1}{2}\text{at}^2=\text{u}+\frac{1}{2}\text{a}(1)^2=\frac{1}{2}\text{a}$
Total distance covered by coin is given by $=1.8+\frac{1}{2}\text{a}=\frac{1}{2}\text{g}$$\Rightarrow1.8+\frac{\text{a}}{2}=\frac{9.8}{2}=4.9$
$\Rightarrow\text{a}=6.2\text{m/s}^2=6.2\times3.28=20.34\text{ft/s}^2.$

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