Velocity & Distance Question

[stealth]

Can someone answer this question for me? You do not need to show the work, just want to compare my answers with yours. I know its pretty simple, but I am having a brain fart.

A package is moving up in an elevator at a steady speed of 3m/s. The package falls off and strikes the bottom of the elevator shaft in 6 seconds.
a) Find the time taken for the package to reach its maximum height?
b) How far above the bottom of the shaft was the package when it fell off?
c) How far above the bottom of the shaft was the package 0.25 seconds after falling off of the elevator?

[msommers]

http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/applications/velocity.html

[stealth]

Originally posted by msommers
http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/applications/velocity.html

I know how to work it out. I want verification my answers are correct.

[msommers]

:rofl:

[inline6turbo]

I just took a stab at it, haven't done this in 6 years and can't remember the equations but I'd like to know what you got too haha


a) Find the time taken for the package to reach its maximum height?

t = 0.31s

b) How far above the bottom of the shaft was the package when it fell off?

height = 158.3m

c) How far above the bottom of the shaft was the package 0.25 seconds after falling off of the elevator?

height = 158.74m


Let me know if I failed haha.

[inline6turbo]

I'm pretty sure I got that wrong now that I think about it, maybe I'll have a stab at it later. I need to refresh my memory haha

[stealth]

Thanks for trying, it looks like you have it right for the most part.

The part that I am questioning is part c for me.

This is what I have...

u= 3m/s
a= -9.81 m/s^2
t= ?
v= 0

a)

v= u+at
0= 3 + (-9.81)(t)
0.3058 s = t

b)

t= 6s - (0.3058s X 2)
t= 5.3884 s

s= ut+1/2at^2
s= (3)(5.3884) + (.5)(9.81)(5.3884^2)
s= 158.58 m

c)

s= ut+1/2at^2
s= (3)(0.25) + (.5)(-9.81)(0.25^2)
s= 0.4434 m

158.58 m + 0.4434 m = 159.02 m

But for c, I think I have to subtract 158.58 from .4434 to get the distance since it has already falling .4434 m, not gained that in height...

:nut:

[inline6turbo]

No the package has gained that in height, as it's rising for ~0.31s before coming to a stop and they're asking for it's height at 0.25s

[stealth]

Originally posted by inline6turbo
No the package has gained that in height, as it's rising for ~0.31s before coming to a stop and they're asking for it's height at 0.25s

"How far above the bottom of the shaft was the package 0.25 seconds after falling off of the elevator?"

But are they not asking for the height .25s after it has dropped from its maximum height?

[Hakkola]

Originally posted by stealth


But are they not asking for the height .25s after it has dropped from its maximum height?

Yes

[inline6turbo]

I read it as they're asking what the height is .25 after falling off the elevator, and at .25 it hasn't reached it's highest point yet.

Up to you how you want to interpret it.

[stealth]

Originally posted by Hakkola


Yes

So would it not be subtracted from the 158m?

inline6turbo, thats the thing I am not sure how to interpret it lol. One way will give me the right answer and the other way the wrong answer, which is right, who knows?

[Hakkola]

Originally posted by inline6turbo
I read it as they're asking what the height is .25 after falling off the elevator, and at .25 it hasn't reached it's highest point yet.

Up to you how you want to interpret it.

It isn't up to interpretation, you're saying two different things here. The first part of your sentence is correct.

If the height is xxx.x when it falls off, it's asking what the altitude is .25s afterwards.

[inline6turbo]

Originally posted by Hakkola

If the height is xxx.x when it falls off, it's asking what the altitude is .25s afterwards.

Which would be higher then when it fell off. Yes?

But not as high as it will reach eventually. Yes?

I'm not sure how I was saying two different things.

Perhaps a drawing is in order lol

[stealth]

Perhaps you are thinking that when it falls off it is continuing at 3m/s so it will gain some height until it reaches 0m/s, where I am thinking as soon as it falls off of the elevator it has reached 0m/s and starts to fall, correct?

[Hakkola]

Originally posted by inline6turbo


Which would be higher then when it fell off. Yes?

But not as high as it will reach eventually. Yes?


No and no.


Originally posted by stealth
... where I am thinking as soon as it falls off of the elevator it has reached 0m/s and starts to fall, correct?

Correct. Otherwise the term word 'falling' likely wouldn't be used.

b) How far above the bottom of the shaft was the package when it fell off?

This answer is derived from: a) Find the time taken for the package to reach its maximum height?

But then again this is a command not a question, so if this is verbatim from an instructors handout we can't be sure of his handle on grammar. Try reading that like it's a question, (Joking aside, I'm right and don't doubt it).

So logically we can figure out that it begins to fall immediately.

Sorry for being blunt, just tired and too lazy to use more words than necessary.

Taking momentum into account would make the correct answer impossible to ascertain since we would need to know the weight of the package and its aerodynamic qualities etc which are not given.

[stealth]

So I should be subtracting instead of this:

c)

s= ut+1/2at^2
s= (3)(0.25) + (.5)(-9.81)(0.25^2)
s= 0.4434 m

158.58 m + 0.4434 m = 159.02 m

Correct way is: 158.58 - .4434 = 158.1366

[inline6turbo]

Hakkola tou're wrong. This is a basic physics question not an engineering aerodynamics question.

The package is travelling upwards initially. Why else would they give u it's initially speed? This isn't a horizontal distance equation where weight matters.

Using your logic we have it's time, acceleration and initial speed of 0 (assuming it stops moving up the instant it leaves the platform which is pushing it)

If youre pushing a ball in your hand upwards at a given speed and you release the ball it will continue to go up. Same situation here

I'm sure that's what they're testing u for here, to make sure u realize that. That's why they gave u the small time of .25s. Because for .31s it's still going up so your height is increasing.

When they use the term falling they are from the perspective that they are on the eleator when it happens. From that view point it would appears it initially is falling because you're moving at a constant speed upwardsand the package is now being forced to slow down.

[Strider]

Originally posted by Hakkola

No and no.




Originally posted by inline6turbo
Hakkola tou're wrong. This is a basic physics question not an engineering aerodynamics question.

The package is travelling upwards initially. Why else would they give u it's initially speed? This isn't a horizontal distance equation where weight matters.

Using your logic we have it's time, acceleration and initial speed of 0 (assuming it stops moving up the instant it leaves the platform which is pushing it)

If youre pushing a ball in your hand upwards at a given speed and you release the ball it will continue to go up. Same situation here

I'm sure that's what they're testing u for here, to make sure u realize that. That's why they gave u the small time of .25s. Because for .31s it's still going up so your height is increasing.

When they use the term falling they are from the perspective that they are on the eleator when it happens. From that view point it would appears it initially is falling because you're moving at a constant speed upwardsand the package is now being forced to slow down.

I'm going with Renee on this one
I haven't gone through her math, but I agree with her logic.

Package velocity is not 0 when is "falls" off the elevator, it's 3m/s up.
Package weight does not affect acceleration.
Aerodynamic drag is assumed negligible.

[stealth]

Originally posted by inline6turbo
Hakkola tou're wrong. This is a basic physics question not an engineering aerodynamics question.

The package is travelling upwards initially. Why else would they give u it's initially speed? This isn't a horizontal distance equation where weight matters.

Using your logic we have it's time, acceleration and initial speed of 0 (assuming it stops moving up the instant it leaves the platform which is pushing it)

If youre pushing a ball in your hand upwards at a given speed and you release the ball it will continue to go up. Same situation here

I'm sure that's what they're testing u for here, to make sure u realize that. That's why they gave u the small time of .25s. Because for .31s it's still going up so your height is increasing.

When they use the term falling they are from the perspective that they are on the eleator when it happens. From that view point it would appears it initially is falling because you're moving at a constant speed upwardsand the package is now being forced to slow down.

I think you are right, because like you stated the force moving 3m/s is removed which means it still travels up until the force of gravity forces it reach a velocity of 0m/s.

[dirtsniffer]

yea, when it 'falls off' the elevator its still moving (briefly) upwards at 3 m/s.

why dont you just go over things with friends from class like everyone else?

[stealth]

I am not in a class.

[Hakkola]

The package doesn't accelerate ever, it goes from 0 to 3m/s immediately. Wouldn't we need the weight of the package to take momentum into account to calculate how much higher it goes when it falls off the elevator? It doesn't matter if it's travelling vertically, it still has momentum.

Yes, you would assume that the package would continue to travel upwards after falling of briefly, but you would also expect it to accelerate from 0 to 3m/s per second, not have it occur instantly.

I know a tennis ball and a bowling ball given the same aerodynamics, when dropped will hit the ground at the same time, but does an object that weighs 10 kg vs one that weighs ounces slow as quickly when moving upwards at 3m/s? I'm thinking that momentum should have an effect on how quickly gravity slows it down and we don't have the weight of the object so we can't calculate this. If that isn't the case then I'm wrong.

[inline6turbo]

You're definitely over thinking this. In simple physics, like grade 11/12 aerodynamics and air resistance are treated as negligible. In this equation weight isnt being accounted for

[Hakkola]

Originally posted by inline6turbo
You're definitely over thinking this. In simple physics, like grade 11/12 aerodynamics and air resistance are treated as negligible. In this equation weight isnt being accounted for

Haha, fair enough. :thumbsup: