A dropped stone falls past a window, determine it's initial height

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Homework Help Overview

The discussion revolves around a kinematics problem involving a stone that falls past a window, with the goal of determining the initial height from which it fell above the window. The problem includes the time taken to pass the window and its height.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the initial velocity at the top of the window using kinematic equations. There is confusion regarding the signs used for acceleration and displacement, particularly whether gravity should be treated as negative or positive. Some participants question the reasoning behind adding the calculated distance to the height of the window.

Discussion Status

There is an ongoing exploration of the calculations and assumptions regarding the direction of acceleration due to gravity. Some participants have provided alternative calculations and raised questions about the consistency of sign conventions in their equations.

Contextual Notes

Participants are grappling with the implications of their chosen coordinate system and how it affects their calculations. There is a lack of consensus on the correct approach to the signs in the equations, which is contributing to the confusion in the problem-solving process.

chickenonrice
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Homework Statement


A falling stone takes 0.33 s to travel past a window 2.2 m tall.
upload_2014-9-30_14-18-16.png

From what height above the top of the window did the stone fall?
Express your answer using two significant figures.

Homework Equations


Kinematic Equations ~
X= Vi (time) + .5gt^2

Vf^2 = Vi^2 + 2gx

The Attempt at a Solution



First I determined what the initial velocity was at the top of the window:

X = Vi (time) + .5gt^2
2.2 = Vi (.33) + .5 (-9.8) (.33)^2
Vi = 8.28 m/s

Then I used this velocity as my final velocity in this equation:[/B]

Vf^2 = Vi^2 + 2gx
8.28^2 = 0^2 + 2 (-9.8)x
x = -3.5meters

When I first inputted 3.5 meters into my online homework, it came out to be incorrect. I randomly decided to add -3.5 meters to 2.2 meters and the answer: 1.3 Meters, came out to be correct.

I don't understand why I am adding -3.5 to 2.2.

Thank you in advance.
 
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chickenonrice said:

Homework Statement


A falling stone takes 0.33 s to travel past a window 2.2 m tall.
View attachment 73842
From what height above the top of the window did the stone fall?
Express your answer using two significant figures.

Homework Equations


Kinematic Equations ~
X= Vi (time) + .5gt^2

Vf^2 = Vi^2 + 2gx

The Attempt at a Solution



First I determined what the initial velocity was at the top of the window:

X = Vi (time) + .5gt^2
2.2 = Vi (.33) + .5 (-9.8) (.33)^2
Vi = 8.28 m/s

Then I used this velocity as my final velocity in this equation:[/B]

Vf^2 = Vi^2 + 2gx
8.28^2 = 0^2 + 2 (-9.8)x
x = -3.5meters

When I first inputted 3.5 meters into my online homework, it came out to be incorrect. I randomly decided to add -3.5 meters to 2.2 meters and the answer: 1.3 Meters, came out to be correct.

I don't understand why I am adding -3.5 to 2.2.

Thank you in advance.
Check your calculation of Vi; I'm seeing a lower value than 8.28 m/s. In particular, make sure the sign that you're ascribing to the acceleration constant matches your choice of coordinate system that the rest of the equation is assuming.
 
I thought the acceleration due to gravity should always remain negative? With a positive acceleration, the Vi changes to 5.049.

By using the positive acceleration, I got 1.3 Meters exactly!

How exactly do you decide when to use a positive or negative acceleration due to gravity?
 
You chose the downward direction positive when you calculated with positive displacement ##\Delta x =2.2 m ##. If downward is positive, the downward gravitational acceleration is also positive.

ehild
 
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