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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.

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.

Grade:12

5 Answers

Pavan kumar
18 Points
14 years ago

Let acceleration of lift be a.

acceleration of ball with respect to the lift = acceleration of ball with respesct to ground - acceleration of the lift

Or, aball(lift) = g - a = 9.8 m/s^2 - a = 32 ft./s^2 - a [ 9.8m = 32 ft. ]

displacement s = 6 ft.

time taken t = 1 s.

initial velocity u = 0

s = ut + 1/2 at^2 = 0*1 + 1/2 (32-a)*1*1 = 1/2 (32-a)

Therefore, 6 = 1/2 (32-a)

Hence a = 20 ft./s^2

Ashish Ranjan
43 Points
7 years ago
The above answer for the question  ‘‘An elevator is descending ...’‘ is absolutely correct i just wanted to clear some concepts . That is,
Here the term
“acceleration of coin with respect to the lift “ is used beacuse the observer is present in the lift,so the accleration of coin that he would observe is “the accleration of coin w.r.t lift , if the observer is present outside the lift , we would simply take into account the “ACTUAL ACCLERATION OF THE COIN”.
Second Thing that i want to conclude that -a person in the elevator drops a coin at the moment the elevator starts.that means the initial velocity of the coin would also be 0.So, Don’t confuse with this term  - The coin is dropped at the instant when evrything is at standstill . 
 Acceleration = (2d/t^2), = 6.142m/sec^2. 6ft. is 1.829m., so distance elevator has drolled in the 1 sec = (4.9 - 1.829) = 3.071m.  Height of fall in 1s = 1/2 (t^2 x g), = 4.9m. Thsi should be the answer.. a=6.14m/sec^2  3.07=0+1/2*a*1  S=ut+1/2at^2 (where a is the acceleration of elevator)  So substituting this S again we have  but the distance to floor was only 1.82m(6ft) that means the elevator moved 4.9-1.82=3.07m in that one second.  S=0x1+1/2x9.8x1=4.9 m  S=ut+1/2gt^2  the coin drops after one sec that means distance covered by coin 
Dweepayan Biswal
40 Points
6 years ago
20 check out hc verma.?........................................................?...........?.?..................?.................................................?
navya
34 Points
4 years ago
for elevator and coin u=0
acceleration of the coin w.r.t to earth – acceleeration of the lift is basicaly
= g-a ( it is coin -earth)
= 9.8- a
 
height of the coin from the elevator is 6ft = 1.8 m aprox 
t =1s
s= ut+1/2at2
1.8 = 0t = ½ (9.8 -a) 12
= 6.2 m/s2
Diya Biju
15 Points
4 years ago
S=6ft
g=32ft/s^2
 t=1 s
u=0
So, s=ut+1/2at^2
6=1/2×a
a=12ft/s^2
(By the relative velocity concept)
Acceleration  of the elevator=32_12=20ft/s^2
 

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