Relative Motion - Elevator Problem

[LHC_23]

Homework Statement

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.

Homework Equations

x = 0.5*g*t²

x = ut + 0.5*g*t²

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 ?[/B]

 

[haruspex]

What is the initial speed of the coin?

 

[BvU]

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 ?

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

Show your workings and we'll guide you

 

[wabbit]

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

 

[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 ?

 

[haruspex]

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 ?

Start by answering my question in post#2.

 

[LHC_23]

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

 

[haruspex]

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

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

 

[LHC_23]

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

 

[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

 

[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 ?

 

[haruspex]

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

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

 

[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 ?

 

[haruspex]

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.

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.]

 

[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.]

 

[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

 

[haruspex]

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

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

 

[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.

 

[haruspex]

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.

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.