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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.

```
11 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
```
11 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
```
4 years ago
```							20 check out hc verma.?........................................................?...........?.?..................?.................................................?
```
3 years ago
```							for elevator and coin u=0acceleration 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 =1ss= ut+1/2at21.8 = 0t = ½ (9.8 -a) 12= 6.2 m/s2
```
2 years ago
```							S=6ftg=32ft/s^2 t=1 su=0So, s=ut+1/2at^26=1/2×aa=12ft/s^2(By the relative velocity concept)Acceleration  of the elevator=32_12=20ft/s^2
```
one year ago
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