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.. Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Solution:

For elevator and coin : Initial velocity=u=0m/s

As the elevator descends  with uniform acceleration "a" the coin has to move more distance than 1.8cm to strike floor.

time taken =t=1 sec

Scoin=Sc=ut+1/2at²

Sc= 0+1/2at²
=1/2 xg x1²
Sc=g/2
Se=ut+1/2 at²
=0+1/2 a(1)²
Se=a/2

Total distance travelled by coin is given by :

1.8 +1/2 a =1/2g
1.8 +a/2 =g/2
1.8+a/2=9.8/2
a/2=4.9 -1.8=3.1m
a=6.2 m/s2
a= 6.2 x 3.28 = 20.34 ft/s²

Answer 2

HEY!!

_____________________________

▶Let the acceleration of the following lift be =
ft/s^2

▶The coin is dropped, it will fall freely because of gravity resulting the starting from rest with acceleration g.

▶Both elevator and coin will go at downward direction.

▶Acceleration of coin (floor) = (g-a) ft/s^2

h= 1/2 (g-a)t^2

6= 1/2 (32-a)1^2

▶▶a = 20 ft/s^2