question_answer

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 (\[\text{g}=\text{32 ft}/{{\text{s}}^{\text{2}}}\]).

Answer:

Let\[x\]in the distance travelled by an elevator as the coin reaches the floor of the elevator. \[\therefore \] total distance travelled by the coin in 1 second \[6+x=\frac{1}{2}\times g\times {{t}^{2}}\Rightarrow 6+x=\frac{1}{2}\times 32\times 1\] \[=16\Rightarrow x=16-6=10m\,\,again\,\,x=\frac{1}{2}a{{t}^{2}}\] \[1=\frac{1}{2}\times a\times 1\Rightarrow a=20ft/{{s}^{2}}\]