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 downward with acceleration $a^{\prime }$ ( say)

The coin has to move more distance than $1.8m$ to strike the floor.

Time taken $t=1sec$

$s_c=ut+\frac{1}{2}{a}^{\prime }{t}^2=0+1/2g(1{)}^2=1/2g$

$s_e=ut+\frac{1}{2}a{t}^2=u+1/2g(1{)}^2=1/2a$

Total distance covered by coin is given by $=1.8+1/2a=1/2g$

$\Rightarrow 1.8+a/2=9.8/2=4.9$

$\Rightarrow a=6.2m/s^2=6.2\times 3.28=20.34ft/s^2$.