# How Fast Will the Flower Pot Hit the Ground?

• tomboi03
In summary: And then use that velocity in the original equation for the speed at the bottom of the window to find the speed at the ground.In summary, the known variables are t, Lw, and g. To find the velocity vground of the pot as it hits the ground, you can use the equation Vb = Lw/t + (g*t)/2 and express the answer in terms of hb, Lw, t, vb, and g. Alternatively, you can use the accelerated motion formulas to find the time to fall the height h and the velocity after that time, and then use that velocity in the original equation for Vb to find the speed at the ground.
tomboi03
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 Lw. 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).

So the known variables are... t, Lw, and g.

If the bottom of your window is a height h above the ground, what is the velocity vground of the pot as it hits the ground? You may introduce the new variable vb , the speed at the bottom of the window, defined by

Vb=Lw/t+(g*t)/2.

Express your answer in terms of some or all of the variables hb, Lw, t, vb, and g.

I have these equations already... but i don't know where to go about from here...

hb=vb+1/2(gt2)
the equation given...
vg=vb+g*t
vb=1/2*g*t2-hb

I keep missing one variable... and I'm getting frustrated... :'(

Thank You,
tomboi03

Okay, so you know the speed at the bottom of the window, h above the ground. You just have to find the speed after it falls that distance h.

"hb=vb+1/2(gt2)" doesn't make sense - it says a distance equals a velocity.
"vg=vb+g*t" would be correct if t was the time to fall the height h, but it is not the same as the t in the question.
"vb=1/2*g*t2-hb" has a distance subtracted from a speed, which is impossible because the units clash.

What you really need here is an accelerated motion formula with initial and final velocities, distance, but not time. If you can locate one on your list, it will solve the question instantly. Failing that, you must use a distance formula like
d = Vot + .5at^2 to find the time to fall the height h. Then use a velocity formula like
V = Vo + at to find the velocity after that time.

I understand your frustration and I am here to help you out. Let's break down the problem and see if we can come up with a solution.

First, let's define our variables:
t = time the pot is visible
Lw = vertical length of your window
g = acceleration due to gravity
hb = height of the bottom of the window above the ground
vb = speed at the bottom of the window
vg = velocity of the pot as it hits the ground

We can use the equation hb = vb + 1/2(gt^2) to solve for vb. We know that at the bottom of the window, the velocity is 0, so we can rewrite the equation as:
hb = 0 + 1/2(gt^2)
Solving for vb, we get:
vb = (2hb/g)^0.5

Now, we can use the equation vg = vb + gt to solve for vg. Substituting our value for vb, we get:
vg = (2hb/g)^0.5 + gt

Since we know that the pot was dropped from the floor above, we can assume that the height of the bottom of the window (hb) is equal to the height of the floor above. So we can rewrite the equation as:
vg = (2(hb+Lw)/g)^0.5 + gt

Now, let's substitute the equation given in the problem for vb:
vb = Lw/t + (g*t)/2

Substituting this into our equation for vg, we get:
vg = (2(hb+Lw)/g)^0.5 + gt = (2(hb+Lw)/g)^0.5 + (Lw/t + (g*t)/2)t
Simplifying this equation, we get:
vg = (2(hb+Lw)/g)^0.5 + Lw + (g*t^2)/2

So, the velocity of the pot as it hits the ground (vg) can be expressed in terms of the variables hb, Lw, t, g. I hope this helps you in solving the problem. Good luck!

## 1. How does the speed of an object falling past a window affect the perception of the person inside?

The speed of an object falling past a window can affect the perception of the person inside in a few ways. If the object is falling at a slow speed, the person may be able to clearly see and track its movement. However, if the object is falling at a high speed, the person may only catch a glimpse of it as it quickly passes by.

## 2. Does the size or shape of the object falling past a window impact its perceived speed?

Yes, the size and shape of an object can impact its perceived speed when falling past a window. Generally, larger and more elongated objects will appear to fall faster than smaller and rounder objects, even if they are falling at the same speed.

## 3. How does the distance from the window affect the perception of an object falling past it?

The distance from the window can greatly impact the perception of an object falling past it. The closer the object is to the window, the more likely it is that the person inside will be able to accurately perceive its speed and movement. However, if the object is far away from the window, it may appear to be moving slower or even seem stationary.

## 4. Can the angle at which an object falls past a window affect its perceived speed?

Yes, the angle at which an object falls past a window can affect its perceived speed. If an object is falling directly perpendicular to the window, it will appear to be falling at its true speed. However, if the object is falling at an angle, it may appear to be falling slower or faster depending on the angle.

## 5. How does the environment outside the window impact the perception of an object falling past it?

The environment outside the window can also impact the perception of an object falling past it. If the background is plain and uniform, the object's movement may be easier to track. However, if the background is cluttered or constantly changing, it may be more difficult for the person to accurately perceive the object's speed and movement.

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