This problem definitely tripped me up and it took me quite a while to understand what the question was actually asking. But I figured it out! Here's an in depth explanation on how to do the problem.
Firstly, it is important to note that the 0.420s is describing the time it took the flowerpot to fall JUST past the window, which is 1.9 m tall. Let me explain how to set up the coordinate system:
Set your positive y-axis downwards instead of upwards. This will prevent you from having to do potentially error-provoking conversions later in the problem. Your positive x-axis should be in the "normal" direction towards the right (although this problem does not actually need an x-axis).
Once you have set up your coordinate system, you can plot your windowsill and window. Below the x-axis, put a dot or line and label it with a variable. This is to signify the top of the window so I will be using W when describing this point. Next to this, you should write 1.9m so you know exactly what your point is describing.
Below this point, you should put another dot or line and label it with a different variable. This is to signify the windowsill from which the flowerpot fell so I will be using F to describe this point.
The space between points W and F is the distance you are looking for in the question. I have labeled this as H.
You are now ready to begin your calculations!
First, you need to find the velocity of the flowerpot as it passes point W (the top of the window and 1.9m off of the ground). To do this, we will use the following equation:
y = yo + voy*t + .5*a*t^2
where y is your ending location, yo is your initial location, voy is your initial velocity, and t is the time the particle (the flowerpot) is moving. In this problem, we are disregarding point F and the distance H.
Substitute in your known values.
(0.0 m/s) = (1.9 m/s) + voy*(0.420s) + .5*(-9.8m/s^2)*(0.420s)^2
It is important to note that your yo is 1.9 m/s. Remember that the pot will pass the top of the window before it reaches y = 0.0 m. Furthermore, remember to put the gravitational acceleration as a negative quantity. A positive quantity would indicate that you were throwing an object into the air instead of the object being in free fall as should be the case in this problem.
Solve for voy.
voy = - 2.465809524 m/s
This quantity makes sense being negative. Remember that velocity is a vector and has both a scalar quantity and a direction. The object is FALLING towards the negative direction and should be a negative quantity.
Now we can solve for distance H. To do this, we will use the following equation:
v^2 = voy^2 + 2*a*(y-yo)
which can be rewritten as:
v^2 = voy^2 + 2*a*d
where d is the distance between the two points you are looking at; in our case F and W.
Before you do any substitutions, think about your diagram you drew out previously. You do not want to make any name changes to quantities if not necessary. Now, when thinking about voy and v, you should consider a typical x-y graph. Now, as y increases, the viewer will run into points below their destination point before reaching this point. In other words, you read your graph from lower quantities to higher quantities, from the negative direction to the positive. You make -2.465809524 m/s your initial velocity because when trying to get a distance, you typically take the most negative quantity as your initial point and the most positive quantity as your final point. Otherwise, you would get a negative distance after calculations. You are looking for the magnitude of distance H, not its vector.
Now substitute known quantities.
(0.0 m/s)^2 = (-2.465809524 m/s)^2 + 2*(-9.8m/s^2)*(Hm)
Again, remember that gravitational acceleration should be negative. Now solve for H.
H = 0.3102151331 m
With significant figures, H is equal to 0.31 m.
Yay. :)
