Solving for the Distance of a Falling Flower Pot

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In summary, the problem involves a flower pot falling from a windowsill and passing a window below in 0.42 seconds. The window is 1.9m high and the question is asking for the distance between the top of the lower window and the windowsill. The solution involves using the equations of motion and conservation of energy, with the final answer being 0.31m.
  • #1
spj1
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1. A flower pot falls out of a window and past the window below (ignore air resistance). It takes the post .42 s to pass this window, which is 1.9m high. How far is the top of the window below the windowsill from which the flowerpot fell?
answer: .31 m




2.
v = v0 + at
x = x0 + v0t + 1/2at^2
v^2 = v0^2 + 2a(x - x0)




3. The Attempt at a Solution
Known facts:
t = .42 s
bottom window height = 1.8
a = -g
x0 = 0 (if the origin is the top windowsill)
v0 = ?
x = ?

x = 0 + v0(.42) + 1/2(-9.8)(.42)^2

So I need to find the initial velocity so I can get the total distance and subtract it from the 1.9. If I had the total distance down, then I could get it easy, but I'm stumped from here since I'm trying to find the total distance. I've tried using v0=0 and that didn't work. I also tried setting the v = 0 (thinking I could do that sense the problem ends with that bottom window). Every route I go it seems I'm still left with two unknowns or I end with a total distance of .86 which isn't even longer than the bottom window.


Any hints on which equations to put together?
 
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  • #2
hi there:

By conservation of energy:
init=final
mgh=1/2mv^2
v^2=2gh
That means that he reach the top of the window with v velocity.
So, we can rewrite the equation of motion as:
(v and g minus because the referential)
x=xi-sqrt(2gh)t-1/2gt^2
x-xi is given:
x is the bottom of the window
xi is the top of the window
t-ti is given(consider ti=0 in top of window)

h is computed: 1 var with 1 equation...

Right?
 
  • #3
Thanks, but you've went a little over my head. We haven't covered anything but the basic equations. To be honest, it seems like there's something missing in the question. Either the total height needs to be given or the velocity...
 
  • #4
ok, you know by conservation of mechanical energy that:

inicial Energy = Final Energy

well, Energy=Kinectic+Potencial

so, inicial K+inicial P=final K+final P
but
[tex]K=0.5*m*v^{2}[/tex]
[tex]P=m*g*x[/tex]

you can consider that, in first step, the x=0 in the windowsill, and the inicial velocity=0,and -xf the distance from windowsill to the top of the window below,
so:
[tex]m*g*-x_{f}=0.5*m*v_{f}^{2}[/tex]
compute in order to v and you get the velocity in the top of the window below.

Now, we know the velocity in the moment the plant reach the window top as function of h(the heigh between the windowstill and the window top.

so, we make the second step: find the velocity needed in the window top so that the plant need 0.48s to reach the end.

to do that, you use the equation of movement, as you have the xi, the t, and the g
so, you can compute v.
x=xi+vo*t-0.5g*t^2
t is given(as you consider t=0 in window top), xi is 1.9m, x=0, g is 9.8. vo is computed

if no, pm me and we get it trough msn or something
 
Last edited:
  • #5
my professor solved this problem the other day with the 3 formulas you stated, very above average difficulty. very
(for intro physics, its overwhelming), maybe littlepigs method is way better, but that takes a little more indepth

ps, he used this problem to introduce the possibility that you can arrange your problem so that T = T - delta T
(delta T being the time that it takes to cross the window, T being the total travel time)
 
Last edited:

Related to Solving for the Distance of a Falling Flower Pot

What is the formula for calculating the distance of a falling flower pot?

The formula for calculating the distance of a falling flower pot is d = 1/2 * g * t^2, where d is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time in seconds.

How do you determine the time in seconds for a falling flower pot?

The time in seconds for a falling flower pot can be determined by using a stopwatch or timer to measure the duration of the fall. Alternatively, you can use the equation t = √(2d/g) to calculate the time based on the distance and acceleration due to gravity.

Can the distance of a falling flower pot be affected by air resistance?

Yes, air resistance can affect the distance of a falling flower pot. The presence of air resistance will cause the pot to reach a terminal velocity, which is a constant speed at which the force of air resistance equals the force of gravity. This will limit the distance the pot travels during its fall.

What is the significance of the acceleration due to gravity in calculating the distance of a falling flower pot?

The acceleration due to gravity is a constant value that determines the rate at which an object falls towards the ground. It is a crucial factor in calculating the distance of a falling flower pot as it affects the speed and duration of the fall.

Can the distance of a falling flower pot be affected by the height from which it is dropped?

Yes, the height from which a flower pot is dropped will affect the distance it falls. According to the formula d = 1/2 * g * t^2, the distance is directly proportional to the square of the time. Therefore, a greater initial height will result in a longer fall time and a greater distance traveled.

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