Physics Free-Fall Kinematics Flowerpot Problem

[Rockoz]

Homework Statement

A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes the pot 0.420 seconds to pass this window, which is 1.90m high. How far is the top of the window below the windowsill from which the flowerpot fell?

Homework Equations

1) V= Vo + at
2) Y = Yo + VYot + (1/2)at^2
3) V^2 = Vo^2 + 2a(Y - Yo)

The Attempt at a Solution

I tried to solve the problem by setting the origin to be the ground, the height of the window as 1.90m, and the height of the initial position of the flowerpot at the windowsill as some unknown Yo (initial y). I believe the Vyo (initial velocity) should be zero. I then used the time .420 seconds to put into the position equation and tried to solve for Yo.

Y = Yo + VYot + (1/2)at^2
1.90m = Yo + 0(.420s) + (1/2)(-9.8 m/s^2)(.420s^2)
(Doing the algebra)
2.65m = Yo or the height of the windowsill

Next I subtracted 1.90m from 2.76m and found .86m to be the distance between the window and the above windowsill but this is incorrect. The answer is 0.31m by the way but I don't know how to get to that answer.

I believe there is something wrong with the way I am picturing and setting up the problem. Thank you very much to anyone who can help.

 

[supratim1]

Y = height of windowsill from which it fell.
so Y - 1.9 = distance traveled in 0.42 s (= required answer)

 

[ehild]

It takes the pot 0.420 seconds to pass this window, which is 1.90m high.

Read the problem again and try to understand what does the above sentence mean.

ehild

 

[Rockoz]

Wow I realize now that I totally misread the problem. I understand it now but hopefully someone can check if my thinking/work is correct. I also hope this topic will be useful to anyone who comes across it in search later.

I establish the origin at the beginning windowsill and make downwards the positive y-axis to avoid dealing with signs. The flower pot travels a distance H + 1.9m ( 1.9m is the length of the window itself). It travels the 1.9m window in 0.420 seconds, not the entire h+1.9m. Next we try to find the velocity of the pot as it passes the very top of the window which I will call Vi. We then solve for Vi (but I do this "seperately" and not really using the origin i had at the beginning for the position equation).

Y = Yo + Vit + 1/2(a)(t^2)
1.9m = O + Vi(0.420) + 1/2(9.8)(0.420^2)
Do the algebra and solve for Vi
2.4658 m/s = Vi

We can use the velocity squared equation to find the distance H the question asks for using this Vi we found. Here I do use the origin I established in the beginning.

V^2 = Vo^2 + 2a(Y - Yo)
(2.4658^2) = 0^2 + 2(9.8) (H - 0)
Do the algebra and solve for H
H = 0.31m

You could also solve it by finding the time it took to reach the Vi velocity and working with that.

A good thing I learned from this exercise is that I must be more careful about reading problems so that I can understand exactly how to set up the problem.

 

[ehild]

Well done! :)

Just an other approach: Let be the origin at the windowsill from where the pot fell. It reaches the top of the window in t seconds, the bottom in t+0.42 seconds. Height of the window is

(g/2)(t+0.42)² -(g/2)t²=1.9.

Solve for t. The distance of the top of the window from the upper windowsill is (g/2)t².

ehild

 

[dawn1392]

This problem definitely tripped me up and it took me quite a while to understand what the question was actually asking. But I figured it out! Here's an in depth explanation on how to do the problem.

Firstly, it is important to note that the 0.420s is describing the time it took the flowerpot to fall JUST past the window, which is 1.9 m tall. Let me explain how to set up the coordinate system:
Set your positive y-axis downwards instead of upwards. This will prevent you from having to do potentially error-provoking conversions later in the problem. Your positive x-axis should be in the "normal" direction towards the right (although this problem does not actually need an x-axis).
Once you have set up your coordinate system, you can plot your windowsill and window. Below the x-axis, put a dot or line and label it with a variable. This is to signify the top of the window so I will be using W when describing this point. Next to this, you should write 1.9m so you know exactly what your point is describing.
Below this point, you should put another dot or line and label it with a different variable. This is to signify the windowsill from which the flowerpot fell so I will be using F to describe this point.
The space between points W and F is the distance you are looking for in the question. I have labeled this as H.

You are now ready to begin your calculations!

First, you need to find the velocity of the flowerpot as it passes point W (the top of the window and 1.9m off of the ground). To do this, we will use the following equation:

y = yo + voy*t + .5*a*t^2

where y is your ending location, yo is your initial location, voy is your initial velocity, and t is the time the particle (the flowerpot) is moving. In this problem, we are disregarding point F and the distance H.

Substitute in your known values.

(0.0 m/s) = (1.9 m/s) + voy*(0.420s) + .5*(-9.8m/s^2)*(0.420s)^2

It is important to note that your yo is 1.9 m/s. Remember that the pot will pass the top of the window before it reaches y = 0.0 m. Furthermore, remember to put the gravitational acceleration as a negative quantity. A positive quantity would indicate that you were throwing an object into the air instead of the object being in free fall as should be the case in this problem.

Solve for voy.

voy = - 2.465809524 m/s

This quantity makes sense being negative. Remember that velocity is a vector and has both a scalar quantity and a direction. The object is FALLING towards the negative direction and should be a negative quantity.
Now we can solve for distance H. To do this, we will use the following equation:

v^2 = voy^2 + 2*a*(y-yo)

which can be rewritten as:

v^2 = voy^2 + 2*a*d

where d is the distance between the two points you are looking at; in our case F and W.

Before you do any substitutions, think about your diagram you drew out previously. You do not want to make any name changes to quantities if not necessary. Now, when thinking about voy and v, you should consider a typical x-y graph. Now, as y increases, the viewer will run into points below their destination point before reaching this point. In other words, you read your graph from lower quantities to higher quantities, from the negative direction to the positive. You make -2.465809524 m/s your initial velocity because when trying to get a distance, you typically take the most negative quantity as your initial point and the most positive quantity as your final point. Otherwise, you would get a negative distance after calculations. You are looking for the magnitude of distance H, not its vector.

Now substitute known quantities.

(0.0 m/s)^2 = (-2.465809524 m/s)^2 + 2*(-9.8m/s^2)*(Hm)

Again, remember that gravitational acceleration should be negative. Now solve for H.

H = 0.3102151331 m
With significant figures, H is equal to 0.31 m.

Yay. :)

 

[ehild]

Hi down, welcome to PF.

If I understood correctly, your positive y-axis points downward so the flowerpot falls upward. OK, it is difficult to imagine, but you can do.
But 1.9 is length (m) not m/s.
Anyway, you got the solution! Cool!

You can attach a figure, it is easier to understand then a written description of the set-up.

I would set up the coordinate system as in the figure. The positive y-axis points downward - why not? This way the flowerpot looks falling downward, gravity is positive, the velocity is positive.
The origin is at the windowsill from where the pot fall. The position of the pot is y(t). At t=0, y(0)=0, v(0)=0. At the top of the window, y₁=H, t=t₁ and y₂=H+1.9(m), t₂=t₁+0.42 (s) at the bottom.

y=g/2 t²:

y₁=H=g/2 t₁²

y₂=H+1.9=g/2 t₂²

Subtract the first equation from the second one:

1.9=g/2 (t₂²-t₁²)
1.9=4.9 (t₁+t₂)(t₂-t₁)
1.9=4.9*0.42(2t1+0.42) ---> t1=0.2516 s, H=4.9 t₁²=0.31 m.

ehild