# At what height is this flower pot dropped?

1. Jan 28, 2005

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 $$h$$ above the bottom of your window was the flower pot dropped?
Express your answer in terms of $$L_W, t,$$ and $$g$$.

I'm trying to find the position right? so i'm going to use the position formula:
x = x(0) + v(0)t + 1/2at^2
well i know that initial velocity is 0, so...

x = $$L_W + 1/2*g*t^2$$

is this correct?

2. Jan 28, 2005

### Justin Lazear

The initial velocity when the pot is dropped is zero, but the initial velocity when you see it at the top of your window is not. It has already been falling for a period of time before you see it, so it's already gained some velocity.

Your current model for the motion would have the velocity at the top of your window being zero. This is clearly not correct.

Choose your coordinate system so that the point where the pot was dropped is the origin. Then the top of your window is at the point y = h, and the bottom of your window is at the point y = h + L. You don't know the time at each of these, but you know the difference in time between these two points, t. So define the time that the pot reaches the top of your window as ttop, then at the point y = h, time = ttop, and at the point y = h + L, time = ttop + t.

Solve for the unknowns using the equations these two points give you.

By the way, it's generally not a very good idea to denote a particular stretch of time as t. Personally, I'd use $\Delta t$ in this case. This is to avoid confusion in statements like "at the point y = h + L, t = ttop + t", which must be rewritten as I did above. t in this problem is a parameter, so using it as a constant as well is not a good idea.

--J