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

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SUMMARY

The discussion focuses on calculating the initial height from which a stone falls past a window that is 2.2 meters tall, taking 0.33 seconds to pass. The initial velocity (Vi) was calculated as 8.28 m/s using the kinematic equation X = Vi (time) + 0.5gt^2. However, a recalculation with a positive acceleration due to gravity yields a different initial velocity of 5.049 m/s, leading to the correct height of 1.3 meters when combined with the window height. The confusion arises from the sign convention used for gravitational acceleration in the calculations.

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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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