How High Was the Elevator When the Nut Fell?

  • Thread starter Thread starter SS2006
  • Start date Start date
  • Tags Tags
    Elevator
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving an elevator and a nut that falls off while the elevator is moving upwards. The problem requires determining the height of the elevator at the moment the nut fell, given the nut's fall time and the elevator's upward velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various equations of motion, including those for velocity and distance. Some express uncertainty about the values they are using in their calculations, while others suggest checking the signs and directions of the variables involved.

Discussion Status

There is an ongoing exploration of different approaches to the problem, with some participants offering guidance on how to set up the equations correctly. Multiple interpretations of the problem are being considered, and participants are encouraged to share their attempts and where they are encountering difficulties.

Contextual Notes

Some participants mention constraints such as the upcoming test and the need to reinforce their understanding of the concepts involved. There is also a focus on ensuring that the correct initial conditions and signs are used in the equations.

SS2006
Messages
80
Reaction score
0
A nut falls off of an elevator moving upwards at 3 m/s. The nut takes 5 s to hit the ground.

Q) How high was the elevator the instant the nut fell off , the answer should be 107.5 metres

i solved this a few weeks ago, now i just can't remmeber it :( casue the test is in 2 days
 
Physics news on Phys.org
Ooops... Well on the bright side, I find solving a problem more than once, reinforces the concepts I'm learning..

How about posting what you have tried so far and show where you are getting stuck? There are many here, who can help steer you in the right direction.
 
well i just can't reach 107.5 m
ive tried v2=v1+at
and d = v1t + 1/2 at squared, the ones that worked last time, i think I am using the wrong numbers
 
[tex]v = v_0 + at[/tex]
&
[tex]v^2 - v_0 = 2as[/tex]
 
This is pretty straight forward,
simply plug your time (5s) into

[tex]d = - \frac {gt^2} 2 + V_0t + h[/tex]

What would your [itex]V_0[/itex] and h be?
 
Did you draw a diagram of your problem, with arrows indicating which direction velocity is moving, which direction acceleration is moving and positions of object?

Let's see, what does the first equation (v2=v1+at) tell you?
Given an intial velocity v1, and an acceleration a, it tells you an object's velocity after some time t.

But you are interested in height (or distance), so first equation doesn't do very much for you.

If you drew arrows indicating the direction of your variables on your initial diagram, you want to also indicate direction in your equation (up or right as + , down or left as -). Using this concept, make sure you use the appropriate signs + or - , in your 2nd equation.
 

Similar threads

  • · Replies 35 ·
2
Replies
35
Views
6K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 11 ·
Replies
11
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K