Solve the following :

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 $6 \mathrm{tt}$ 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.

Solution:

For coin-lift

$\mathrm{u}_{\mathrm{re}}=0 \mathrm{~m} / \mathrm{s}$

$\mathrm{t}_{\mathrm{re}}=1 \mathrm{sec}$

$\mathrm{s}_{\mathrm{rel}}=6 \mathrm{ft}$

$\mathrm{S}_{\mathrm{rel}}=\mathrm{U}_{\mathrm{rel}} t+\frac{\overline{2}}{{ }^{2}} \mathrm{a}_{\mathrm{rel}} t^{2}$

$6=\stackrel{\frac{1}{2}}{a_{\text {rel }}}(1)^{2}$

$\mathrm{a}_{\mathrm{rel}}=12 \mathrm{ft} / \mathrm{sec}^{2}$

$g-a_{\text {lift }}=12$

$\mathrm{a}_{\text {lift }}=32-12$

$a_{\text {lift }}=20 \mathrm{ft} / \mathrm{sec}^{2}$

 

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