Question

As you look out of your dorm window, a flower pot suddenly falls past...

Answer 1

As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your window is\[L _{w}\]
Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g.
Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward)

From what height above the bottom of your window was the flower pot dropped?

Answer 2

Nope. My solution began to become too convoluted. Looking for a more simple solution -- someone else can probably help you out in the mean time.

Answer 3

can i use \[v ^{2}=v _{0}^{2}+2a(x-x _{0})\] to find the velocity at the bottom of the window?

Answer 4

I disagree with the above approach because that assumes it was dropped directly above the top of the window. Using the same equation,
\[ \Delta y = v_0 t + \frac{1}{2}gt^2\]
substituting the height of the window and solving,
\[ v_0 = \frac{L_w}{t} - \frac{gt}{2} \]
So as the pot was moving at the above speed when it first came into view. Assuming the pot was dropped from rest, we can use
\[ v_f^2 = 2g h\]
to determine that the height above the top of the window that the flowerpot was dropped was
\[ \left( \frac{L_w}{t} - \frac{gt}{2} \right)^2 / 2g \]

Answer 5

EdG: Yes, but not directly, since you only know v_0, a, and either x or x_0 (not both).

Answer 6

Jemurray3, I'm going to try your approach

Answer 7

Well, the full solution is there, so are there parts that you don't understand?

Answer 8

it is not correct though, i think there is a mistake in your calculations but I'm not certain where it is.

Answer 9

i inputed that answer and it came out wrong

Answer 10

It wants the height above the bottom of the window, not the top.

Answer 11

Add L_w to whatever you got.

Answer 12

I think the question is asking for a general equation, so:
\[((\frac{L_w}{t}- \frac{g t}{2})^2/2g)+L_w\]Although the problem asked for the positive velocity to be down, so that's going to make some signs funky. Really, you would need to subtract the length of the window.

Answer 13

No you wouldn't.

Answer 14

All of those signs were assuming down as the positive direction.

Answer 15

you are correct Jemurray3.

Answer 16

thanks for the help both of you, i appreciate your time

Answer 17

No problem.

Answer 18

i have to find the velocity as it hits the ground, i think i have that covered. ill come back if i need more help.

Answer 19

there, thats the velocity as it hits the ground. took me quite some time, but i got it.

Answer 20

Good job.