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

[chickenonrice]

Homework Statement

A falling stone takes 0.33 s to travel past a window 2.2 m tall.

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.

 

[gneill]

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.

 

[chickenonrice]

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?

 

[ehild]

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

 

[HalfLostAndSearching]

I would first like to point out that the initial height of the dropped stone cannot be determined with complete accuracy based on the given information. This is because the time taken to travel past the window is only given to two significant figures, meaning that the uncertainty in this value is at least 0.01 s. This uncertainty will propagate through to the calculated initial height and result in a final answer that is also only accurate to two significant figures.

With that being said, let's look at the solution provided. The initial velocity calculated (8.28 m/s) is the velocity at the top of the window, which means that the stone must have been dropped from a height greater than 2.2 m. This is why the calculated height of -3.5 m does not make sense, as it would indicate that the stone was dropped from below the window.

Adding -3.5 m to 2.2 m is not a valid approach to solving this problem. Instead, we can use the kinematic equation Vf^2 = Vi^2 + 2gx to solve for the initial height (x) using the initial velocity (Vi) and time (t) values that we have already calculated.

Vf^2 = Vi^2 + 2gx
(0 m/s)^2 = (8.28 m/s)^2 + 2(-9.8 m/s^2)x
x = -3.41 m

This calculated initial height of -3.41 m indicates that the stone was dropped from a height greater than 2.2 m, which is consistent with our initial assumption. However, as mentioned earlier, the uncertainty in the given time value means that the calculated initial height is only accurate to two significant figures. Therefore, the final answer should be rounded to -3.4 m.

In summary, the initial height of the dropped stone cannot be determined with complete accuracy based on the given information. However, using the given time and the calculated initial velocity, we can determine that the stone was dropped from a height of approximately -3.4 m above the top of the window.