At what height is this flower pot dropped?

[ProBasket]

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?

 

[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 t_top, then at the point y = h, time = t_top, and at the point y = h + L, time = t_top + 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 = t_top + 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

 

[wais]

Yes, your approach is correct. However, there is a small mistake in your formula. The initial position, x(0), should be the height from which the flower pot was dropped, not the height of your window. So the correct formula would be:

x = h + 1/2*g*t^2

We can rearrange this formula to solve for h:

h = x - 1/2*g*t^2

Since x is the vertical length of your window, L_w, we can substitute it into the formula to get the final answer:

h = L_w - 1/2*g*t^2

Therefore, the height from which the flower pot was dropped is L_w - 1/2*g*t^2. This formula takes into account the acceleration due to gravity, the time the pot was visible, and the height of your window.