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

AI Thread Summary
The discussion revolves around calculating the initial height from which a stone fell before passing a 2.2 m tall window in 0.33 seconds. Initial velocity was calculated as 8.28 m/s, but there was confusion regarding the use of negative acceleration in the kinematic equations. A participant pointed out that using a positive acceleration due to gravity led to a correct answer of 1.3 meters when combined with the window height. The key takeaway is the importance of consistently defining the coordinate system, particularly the direction of acceleration, to avoid errors in calculations. Understanding when to apply positive or negative values for gravity is crucial in solving such physics problems accurately.
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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