At what height is this flower pot dropped?

  • Thread starter ProBasket
  • Start date
  • Tags
    Height
In summary, the conversation discusses the motion of a flower pot that falls past a dorm window. The pot is dropped from a height h above the bottom of the window, and the vertical length of the window is L_W. The acceleration due to gravity is represented by g and the time the pot is visible is represented by t. The position formula x = L_W + 1/2*g*t^2 is used to find the position of the pot, but it is noted that the initial velocity when the pot is dropped is not zero. It is suggested to define the time the pot reaches the top of the window as ttop and to use the equations at both the top and bottom of the window to solve for the unknowns. It
  • #1
ProBasket
140
0
As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time [tex]t[/tex] , and the vertical length of your window is [tex]L_w[/tex]. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity [tex]g[/tex] .

Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).


From what height [tex]h[/tex] above the bottom of your window was the flower pot dropped?
Express your answer in terms of [tex]L_W, t,[/tex] and [tex]g[/tex].

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 = [tex]L_W + 1/2*g*t^2[/tex]

is this correct?
 
Physics news on Phys.org
  • #2
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 [itex]\Delta t[/itex] 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
 
  • #3


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.
 

1. What is the relationship between the height of a dropped flower pot and its impact?

The height at which a flower pot is dropped can greatly affect its impact upon landing. The higher the drop, the greater the force of impact due to the increased potential energy of the pot.

2. Does the weight or material of the flower pot affect the impact at different heights?

Yes, the weight and material of the flower pot can impact the force of impact at different heights. A heavier or more rigid pot may have a greater impact than a lighter or more flexible pot when dropped from the same height.

3. Is there a point at which the height of a dropped flower pot no longer affects its impact?

No, there is no point at which the height of a dropped flower pot will not affect its impact. However, the difference in impact between small changes in height may become negligible at extremely high heights.

4. How does the surface or environment the flower pot is dropped onto affect its impact at different heights?

The surface or environment that the flower pot is dropped onto can greatly impact its force of impact at different heights. A hard, solid surface may result in a greater impact compared to a soft, cushioned surface.

5. Can the shape of the flower pot affect its impact when dropped from different heights?

Yes, the shape of the flower pot can impact its force of impact when dropped from different heights. A pot with a wider or more rounded base may have a greater impact compared to a pot with a narrower or more pointed base.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
5K
  • Introductory Physics Homework Help
Replies
8
Views
5K
  • Introductory Physics Homework Help
Replies
25
Views
387
  • Introductory Physics Homework Help
Replies
34
Views
597
  • Introductory Physics Homework Help
Replies
1
Views
7K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
253
  • Introductory Physics Homework Help
2
Replies
38
Views
1K
Back
Top