# Homework Help: Relative Motion - Elevator Problem

1. Apr 21, 2015

### LHC_23

1. The problem statement, all variables and given/known data
A lift in which a man is standing is moving upwards with a speed 10 m/s. The man drops a coin from a height of
4.9 m and if g = 9.8 m/s2 - then after what time does the coin reach the floor.

2. Relevant equations
x = 0.5*g*t2

x = ut + 0.5*g*t2

3. The attempt at a solution
I know this is quite a simple problem - my point is, why does an observer inside the elevator observe
the coin touching the floor at t = 1s while the observer on the ground would observe the coin touching
the floor at t = sqrt(2) - 1 s ?

2. Apr 21, 2015

### haruspex

What is the initial speed of the coin?

3. Apr 21, 2015

### BvU

Apparently it isn't that simple, because the answer to your question is : Neither nor !

Show your workings and we'll guide you

4. Apr 21, 2015

### wabbit

Edit: reply removed, was posted at the same time as others, no need for a chorus :)

5. Apr 21, 2015

### LHC_23

Well, here is what I what I thought :

From the reference frame of the elevator - The initial velocity of the coin must be 0 - since they were both moving upwards with 10 m/s. Then I applied the equation
x = 0.5*g*t2 - and plugging in x=4.9 m and g = 9.8 m/s2 - I got t = 1 s.

From the reference frame of the ground - I'm not really sure now about how I got that answer - must have been a mistake I guess. But, could someone tell me how I should approach the problem from the viewpoint of the ground ?

6. Apr 21, 2015

### haruspex

Start by answering my question in post#2.

7. Apr 21, 2015

### LHC_23

@haruspex - From the reference frame of ground, it should be 10 m/s upward.

8. Apr 21, 2015

### haruspex

Ok. If it takes time t to hit the floor of the elevator, how far will it have travelled relative to the ground?

9. Apr 21, 2015

### LHC_23

It would have travelled x = -10t + 4.9t2

Last edited: Apr 21, 2015
10. Apr 21, 2015

### wintjc

You consider the upwards velocity only if you let the coin drop outside the elevator There is no time difference within the elevator for observers inside or out

11. Apr 21, 2015

### LHC_23

@wintjc But then, I don't see the difference between dropping it outside or inside - if I am in the reference frame of the ground, do I ?

12. Apr 21, 2015

### haruspex

Ok, and what is x in terms of the height at which it was dropped, given the elevator's motion.

13. Apr 21, 2015

### LHC_23

I think I didn't really get your question - How do I express x in terms of "the height at which it was dropped, given the elevator's motion" ?

Or rather : I don't know how to go about it - given my understanding of your question.

EDIT : I still don't get how things would work from the ground frame - Could you show me how you would go about solving the problem from the ground frame ?

Last edited: Apr 21, 2015
14. Apr 21, 2015

### haruspex

You are given the height from the elevator floor at which it was dropped, yes? We have set t as the time it takes to hit the floor, and x as the distance it moves (relative to ground) in the process. How far has the elevator floor moved in time t?
[One thing worries me - 4.9 m is a very tall elevator cage.]

15. Apr 21, 2015

### LHC_23

Well, the elevator floor would have moved a distance 10t.

[ :D , I hadn't noticed that - I was just assuming a person (had to be a giant) took a penny out from his pocket and dropped it by mistake - I didn't assume the height of the elevator cage to be 4.9 m - Though I think that would represent a practical situation much better.]

16. Apr 21, 2015

### wintjc

If the floor is the ground and the height off the floor is 4.9 m then use the upwards initial velocity in the equation offset by g. And solve If the floor is the elevator floor then it's just a 4.9 m drop

There us no time difference unless you apply two different scenarios for two observers. Ie doing two different problems. You are making this more complex than it is

17. Apr 21, 2015

### haruspex

My reading of the thread is that LHC understands that, but was puzzled to get different results by the two methods.

18. Apr 21, 2015

### BvU

I 'm following this thread with a sense of puzzlement: 4.9 m high is nonsense for an elevator cage. My reading for this exercise is: the coin falls from 4.9 m high with an initial upward speed of 10 m/s, so it takes longer than 1 s to reach the floor.

But then, an upward speed of 10 m/s at a distance 4.9 m above the floor requires an equally nonsensical acceleration if you start from zero. Perhaps the thing took off from the -5th floor in a deep underground parking garage ?

As an exercise, I find it of miserable quality: 4.9 m, 10 m/s ? And: Dropping things out of an elevator cage is generally impossible and therefore hard to imagine.

19. Apr 21, 2015

### haruspex

I also wondered whether 'floor' meant ground, rather than the floor of the cage. Or maybe 4.9m is a typo for 0.49 m.