# Elevator question

1. Sep 15, 2004

### tony873004

While riding on an elevator descending with a constant speed of 3.4 m/s, you accidentally drop a book from under your arm.
(a) How long does it take for the book to reach the elevator floor, 1.2 m below your arm?
s
(b) What is the book's speed (relative to the earth) when it hits the elevator floor?
m/s

I tried coming up with a new acceleration value by subtracting 3.4 from 9.81,but I got the wrong answers. Then it dawned on my that I can't do it that way. If the elevator were decending 9.8 m/s, objects would still drop to the elevator floor because they're accelerating and the elevator is moving at a constant speed.

So... I'm lost on this one. There's an almost identical problem about a hot air balloon ascending at a constant rate, and someone throwing a camera up to someone in the balloon. But if I understand how to do the first one, I can probably come up with the other.

2. Sep 16, 2004

### tony873004

I figured it out. The fact that the elevator is moving does not affect the acceleration, but its speed does needed to be to the book's velocity to get Earth relative velocity.

3. Sep 16, 2004

### maverick280857

Though you have it figured out now Tony, I still have a few suggestions:

Before the book left your hand, it was traveling upwards with the same velocity as that of the lift (assuming of course that you are not shaking vigourously in the lift or doing something to cause relative motion of the book with respect to the lift ). The instant it left your hand therefore, it had the same instantaneous velocity as that of the lift, directed upwards. However, for all time instants subsequent and until the time it landed on the floor of the lift, it experienced an acceleration of -g vertically downwards.

But then again, this isn't your simple kinematic problem...the lift is moving upwards too so the distance between the book and the floor is being decreased both due to the book's downward motion and the lift's upward motion. This really suggests a trick...relative velocity. Suppose I visualize the lift to be at rest relative to the block. Hence, the block has an instantaneous relative velocity of zero when it leaves the hand. It has an acceleration -g relative to the lift (since accel(lift) = 0 and accel(book) = -g so accel(book/lift) = accel(book)-accel(lift) = -g). With these kinematic variables in place, it covers a distance equal to 1.2 m. You can use 1.2 = (1/2)gt^2 to solve for t.

Hope that helps...

Cheers
Vivek

(PS--Balloon problems are similar :-D)

4. Sep 16, 2004

### tony873004

Actually, I can't figure out the balloon one yet. They're not as similar as I thought.

A hot air balloon has just lifted off and is rising at the constant rate of 2.2 m/s. Suddenly, one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 11.6 m/s. If the passenger is 2.5 m above her friend when the camera is tossed, how high is she when the camera reaches her on its way up?

5. Sep 16, 2004

### maverick280857

Well I'll give you a hint Tony, but I really think that you should try and find the correlation with the elevator problem.

The initial velocity of the camera is 11.6 m/s when its separation with the balloon is 2.5m. The camera's motion is decelerated (under the influence of acceleration due to gravity acting downwards at all times). The balloon is rising at a constant rate of 2.2 m/s so what do you think happens when the camera catches up with the balloon?

6. Sep 16, 2004

### tony873004

The girl in the balloon catches it. But it is not known if she catches it while the camera is at the high point of its trajectory (in which case this would be an easy question), or if it still has upward velocity.

I almost want to solve this numerically on the computer, or use calculus. But I'm sure it's simpler than that. I know I'm just missing something.

7. Sep 16, 2004

### Ba

Make your two kinematics equations equal in distance remembering the initial distance of the balloon, if the camera is not caught at the apex then you will recieve two answers, (this is more probable than the apex) and then you take the smallest times as the question says on the way up.

8. Sep 16, 2004

### NateTG

There is an underlying theme to these questions which is frames of reference.

Regarding the ballon question:
When the camera is thrown, what is it's velocity relative to the friend on the ground?
When the camera is thrown, what is it's velocity relative to the girl in the balloon?
While the camera is in the air, at what rate is it accelerating relative to the friend on the ground?
While the camera is in the air, at what rate is it accelerating relative to the girl in the ballon?

9. Sep 21, 2004

### tony873004

Our teacher gave us a clue, and it's to do it the way Ba suggested.

I got:
camera $$d=11.6 t + 0.5 * -9.81 t^2$$
balloon $$d=2.2 t + 2.5$$

so...
$$11.6 t + 0.5 * -9.81 t^2 = 2.2 t + 2.5$$
That gets rid of the d's in my equations, so now I can solve for t. Then I can plug t into the easy kinmatic equations and get d which is what I'm looking for.

But I've got t on both sides of the equation, and that confuses me .
So I re-wrote it as:
$$11.6t + 0.5 * -9.81 t^2 - 2.2t = 2.5$$
which becomes:
$$9.4 t -4.905 t^2 = 2.5$$

Now is where I get stuck. I love physics and I hate algebra

Did I do it right so far, and if so, what's my next step?

BTW... 0.3191 is the correct answer for t, and 3.202 is the correct answer for d. I got this by solving it numerically, but I wan't to know how to do it the other way (the teacher won't let me bring my desktop computer to the test )

10. Sep 21, 2004

### NateTG

$$9.4 t -4.905 t^2 = 2.5$$
is equivalent to:
$$4.905 t^2-9.4t+2.5=0$$
$$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
so
$$\frac{9.4 \pm \sqrt{(-9.4)^2-4(4.905)(2.5)}}{2(4.905)}$$
I'm too lazy to work out the numbers, but one of the answers is certainly in the right neighborhood. (You should be able to figure out why there are two answers, and why that one is the right answer.)

You should also rember to keep track of sig figs, since this is a physics problem.

11. Sep 21, 2004

### tony873004

Thanks, Nate. I was confusing a and c in my quadratic. Now it works.

As far as the significant figures go, since I had the following figures to begin with:

2.2 m/s
11.6 m/s
2.5 m
9.81 m/s^2

then I assume my answer of 3.202 (which the computer calls correct) has too many significant figures. Would 3.2 be correct? I wish the computer program (webassign) would tell me "correct, but check your significant figures" instead of just "correct".

12. Sep 21, 2004

### NateTG

Yes, 3.2 has the correct number of sig figs.