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

1. Sep 30, 2014

### chickenonrice

1. The problem statement, all variables and given/known data
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?

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

Vf^2 = Vi^2 + 2gx

3. 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:

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.

2. Sep 30, 2014

### Staff: Mentor

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.

3. Sep 30, 2014

### 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?

4. Sep 30, 2014

### 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