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 ft 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 1

For elevator and coin, u = 0

As the elevator descends down ward with acceleration 'a' (say) the coin has to move more distance than 1.8 cm to strick the floor.

Time takes, t = 1 sec

$S_e=ut+\frac{1}{2}a{t}^2$

=$0+\frac{1}{2}g(1{)}^2=1\frac{1}{2}g$

$S_e=ut+\frac{1}{2}a{t}^2$

= $0+\frac{1}{2}a\left(1{)}^2\right)=\frac{1}{2}a$

Total distance covered by coin given by

$1.8\frac{1}{2}a=\frac{1}{2}g$

$\Rightarrow 1.8+\frac{a}{2}=\frac{9.8}{2}$ = 4.9

$\Rightarrow \frac{a}{2}$ = 4.9 - 1.8 = 3.1 m

$\Rightarrow a=6.2m/s^2$

=$6.2\times 3.28=20.34ft/s^2$