How Far Is The Top Of The Window Below The Windowsill From Which The Flowerpot Fell?
Physics Free-Autumn Kinematics Flowerpot Problem
- Thread starter Rockoz
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Homework Argument
A flowerpot falls off a windowsill and falls by the window below. Yous may ignore air resistance. It takes the pot 0.420 seconds to pass this window, which is one.90m high. How far is the superlative of the window beneath the windowsill from which the flowerpot fell?
Homework Equations
i) V= Vo + at
two) Y = Yo + VYot + (1/two)at^2
3) V^2 = Vo^2 + 2a(Y - Yo)
The Endeavour at a Solution
I tried to solve the problem past setting the origin to be the basis, the height of the window every bit i.90m, and the height of the initial position of the flowerpot at the windowsill every bit some unknown Yo (initial y). I believe the Vyo (initial velocity) should be zero. I and so used the time .420 seconds to put into the position equation and tried to solve for Yo.
Y = Yo + VYot + (ane/two)at^two
1.90m = Yo + 0(.420s) + (1/two)(-9.viii g/s^2)(.420s^2)
(Doing the algebra)
2.65m = Yo or the acme of the windowsill
Side by side I subtracted 1.90m from two.76m and found .86m to be the distance between the window and the above windowsill only this is incorrect. The answer is 0.31m by the way just I dont know how to become to that respond.
I believe there is something wrong with the way I am picturing and setting upwards the problem. Thank y'all very much to anyone who tin can help.
Answers and Replies
then Y - 1.9 = distance traveled in 0.42 southward (= required answer)
It takes the pot 0.420 seconds to pass this window, which is 1.90m loftier.
Read the trouble once again and endeavour to understand what does the above sentence hateful.
ehild
I establish the origin at the beginning windowsill and brand 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 unabridged h+1.9m. Next we try to find the velocity of the pot as information technology passes the very top of the window which I will phone call Vi. We then solve for Six (but I exercise this "seperately" and not really using the origin i had at the beginning for the position equation).
Y = Yo + Vit + i/2(a)(t^2)
1.9m = O + Vi(0.420) + ane/two(9.8)(0.420^2)
Practice the algebra and solve for 6
2.4658 m/due south = Half-dozen
We can use the velocity squared equation to find the altitude H the question asks for using this Vi we found. Here I practice apply the origin I established in the beginning.
Five^2 = Vo^2 + 2a(Y - Yo)
(two.4658^ii) = 0^2 + 2(9.8) (H - 0)
Do the algebra and solve for H
H = 0.31m
Y'all could also solve it by finding the time it took to reach the Vi velocity and working with that.
A practiced matter I learned from this exercise is that I must exist more careful nearly reading issues so that I can understand exactly how to set the trouble.
Only an other approach: Let exist the origin at the windowsill from where the pot brutal. It reaches the acme of the window in t seconds, the bottom in t+0.42 seconds. Pinnacle of the window is
(g/2)(t+0.42)ii -(chiliad/two)ttwo=i.9.
Solve for t. The altitude of the top of the window from the upper windowsill is (g/2)t2.
ehild
Firstly, it is of import to note that the 0.420s is describing the time it took the flowerpot to fall JUST by the window, which is 1.9 m tall. Permit me explain how to prepare the coordinate system:
Set your positive y-axis down 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 non actually need an ten-axis).
In one case you have prepare up your coordinate system, y'all can plot your windowsill and window. Below the x-centrality, put a dot or line and label information technology with a variable. This is to signify the top of the window so I will be using Due west when describing this betoken. Next to this, you should write one.9m so you know exactly what your indicate is describing.
Beneath this signal, you should put another dot or line and characterization it with a different variable. This is to signify the windowsill from which the flowerpot cruel so I will exist using F to depict this point.
The space between points W and F is the distance you are looking for in the question. I accept labeled this as H.
Yous are now ready to begin your calculations!
First, you need to find the velocity of the flowerpot as it passes point W (the peak of the window and 1.9m off of the ground). To practise this, we will use the following equation:
y = yo + voy*t + .v*a*t^two
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 altitude H.
Substitute in your known values.
(0.0 m/due south) = (1.9 m/s) + voy*(0.420s) + .5*(-ix.8m/south^ii)*(0.420s)^two
Information technology is important to note that your yo is ane.nine m/s. Recall that the pot will pass the top of the window earlier it reaches y = 0.0 1000. 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 equally should be the example in this problem.
Solve for voy.
voy = - two.465809524 g/due south
This quantity makes sense being negative. Call back 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 nosotros can solve for altitude H. To practice this, nosotros will use the following equation:
5^2 = voy^2 + two*a*(y-yo)
which can be rewritten as:
five^2 = voy^2 + 2*a*d
where d is the altitude betwixt 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 near voy and 5, you should consider a typical x-y graph. Now, as y increases, the viewer will run into points beneath their destination signal before reaching this point. In other words, y'all read your graph from lower quantities to higher quantities, from the negative direction to the positive. You make -2.465809524 chiliad/s your initial velocity because when trying to become a distance, you lot typically take the about negative quantity every bit your initial point and the virtually positive quantity every bit your last point. Otherwise, you would become a negative distance after calculations. Yous are looking for the magnitude of distance H, non its vector.
Now substitute known quantities.
(0.0 thou/s)^two = (-2.465809524 m/s)^two + 2*(-ix.8m/s^ii)*(Hm)
Again, recall that gravitational acceleration should be negative. Now solve for H.
H = 0.3102151331 one thousand
With significant figures, H is equal to 0.31 m.
Yay. :)
If I understood correctly, your positive y axis points downwardly and so the flowerpot falls upwards. OK, it is hard to imagine, simply you can practice.
But one.9 is length (m) not grand/southward.
Anyway, you got the solution! Cool!
You tin attach a effigy, it is easier to understand and then a written clarification of the set-up.
I would set up the coordinate system every bit in the effigy. The positive y axis points downwards - 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, 5(0)=0. At the meridian of the window, yi=H, t=t1 and ytwo=H+1.9(m), t2=tane+0.42 (southward) at the bottom.
y=g/2 t2:
y1=H=one thousand/2 tane 2
y2=H+1.9=g/2 ttwo ii
Subtract the starting time equation from the second one:
one.9=chiliad/ii (t2 2-tone 2)
1.9=4.nine (t1+t2)(t2-t1)
1.ix=4.nine*0.42(2t1+0.42) ---> t1=0.2516 s, H=four.9 t1 2=0.31 m.
ehild
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How Far Is The Top Of The Window Below The Windowsill From Which The Flowerpot Fell?,
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