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YumYum247
 one year ago
Help ME Please!!!!!!!! :)
YumYum247
 one year ago
Help ME Please!!!!!!!! :)

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YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0Hi there please help me!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hello give me one second..

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0ok take your time!!! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well for starters this is a matter of forces on objects.... let me ask you a question.... while the package is in the plane what forces are acting on it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the potential ENERGY the object has is given by what force? lets start there

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0the package is travelling with the velocity of the plane so whatever the seed of the plane is, that's the speed of the package inside the plane and since it's moving in a direction it'll land at an angle with respect to it's reference point( plane)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the FORCES on the object in the plane are ?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0because as the package is dropped from the plane, it'll travel both in the vertical and horizontal axis....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dont overthink it come on

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes yes but the forces?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441073471996:dw

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0potential and kinectic O_o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0those are the energys.... ok one of the FORCES is gravity right? gravity is always acting on everythin.... unless we flying in space right?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0when the package was i nthe plane, the package was also travelling with the plane, but when the package was dropped, it turned into kinectic.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and as you said before we have two axes in which we are travelling vertical and horizontal

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0Oh.....so that's what you meant so 9.81m/sec^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so gravity is the only force in the vertical direction... yes exactley thats g the ACCELERATION due to gravity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what about the force in the horizontal direction? you already mentioned it was on a plane right?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0but there is the potential force of the plane dragging it on the horizontal axis to an angle.

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0i'm referring to the package tho.

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0i think we first need to calculate how far on the horizontal axis the medical supplies are dropped?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0from the campsite.....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0perhaps but I am trying to walk you through how you should think about these things in general

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441073888474:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes how you are thinking about it geometrically is valid dont take what I am saying as you are wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i want to try and walk you through a way of thinking about these things so you can tackle these types of probelms and harder

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so first step is to consider the forces on the package

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well for starters there is gravity acting down on the package through its center of mass

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0is there a step wise procedure to get at these types of problems??????

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but there is also the force of the plane's engines on the package

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont know what level you are in physics so i would say yes considering forces is always a valid starting point for these types

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea :D that was what i was getting at

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if the plane was applying a force to the package then the package wouldnt be flying along with the plane no?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0however once the package is thrown out of the plane.... what forces act on it in a horizontal direction

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0gravitational force!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no im sorry yum kinetic is not a force

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes gravity is still acting on it but that is in the vertical direction

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0yeah i thought about that too late.....:D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you are confusing the two concepts of force and kinetic energy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its alright no worries :D

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0:D sorry pal i'll try not to do it again!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dont worry about it... ok so think really hard what is acting on the package HORIZONTALLY once its been thrown out of the plane

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0gravitational force is acting on the package.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0correct.... and is there anything acting on the package horizontally?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0btw no is a valid answer...

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0would that be gravity as well. because the package is in the free fall, the only thing there is is gravity!!!

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0i think the gravity is making the package distace both vertically and horizontally.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0YES!!! EXACTLEY :D :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0soooooo what does that mean for the horizontal veloctiy?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0the package is covering a certain distance at a constant speed, but on the vertical axis, the package is accelerating downward as the gravitational force pull the package down.

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0so the horizontal component is constant but the vertical component is changing.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so clearly he cant drop it right over the campsite because the constant veloctiy HORIZONTALLY of the package will take it a further distance horizontally

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0during the time in which it is falling

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0so how do we figure out how far ahead of the campsite the medical packages were dropped????

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok brilliant we know the distance between the campsite and the plane (vertical distance)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we know the initial VERTICAL velocity (pre fall) was zero (because the plane was neither climbing nor falling)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441074773233:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and we know the ACCELERATION the package undergoes VERTICALLY while falling

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes so how do we calculate the time of fall?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0ok so to figure out the horizontal displacement, we first need to figure out the time it took to again that distance on the horizontal axis????????

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no in order to figure out the horizontal distance we first need to figure out the time of fall

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in other words the time it takes to cover the vertical distance.... that we know

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you know how to set that equation up?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0let me try first....:)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0Change in vertical displacement = initial velocity X t + 1/2(g X t^2) and solve for time....

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0do you want me to draw for you the equation??????????

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441075219970:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441075283811:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so here we will take the intial position to be zero

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441075275924:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no no dont do that....

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0whyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy!!!?!?!?!?!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you are confusing the initial velocity int he x direction for an initial velocity in the y direction....hold on:

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0ok so i have to figure out the vertical component and horizontal component???

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0i mean do i have to.....???

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0but need the angle for that :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x(t)=x_{0}+v_{i_{x}} t + \frac{1}{2}a t^{2} = v_{i_{x}}\] && \[y(t)=y_{0}+v_{i_{y}} t + \frac{1}{2}a t^{2} =\frac{1}{2}g t^{2} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so we know y(T)=150m (the total fall height)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so that gives us: \[y(T)=150m=\frac{1}{2}gT^{2} \ \ \ \ \rightarrow \ \ \ \ T^{2}=\frac{2}{g}*(150m) \approx 30.6\] \[Therefore \ \ T=\sqrt{30.6} \approx 5.52s\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How did I know the units were seconds.... welll look at the units on the first equation: \[T^{2}=\frac{1}{\frac{m}{s^{2}}}*m=s^{2} \ \ \ \rightarrow \ \ \ T=s\]

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0wait let me label all the given values.... Vertical: Horizontal: g = 9.8m/sec dh =? dv = 150m vh =? vi = 40m/sec t = ? vf = 0m/sec t = ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its a good exercise in unit analysis to check for yourself

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes very good idea at the begining of every problem.... write down your knowns and your unknowns

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0but idk why you applied that formula tho :O

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0wait let me try doing it my wait and see if i get 5.52sec. Hold up, din't go anywhere!!! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which formula? those are the kinetic motion formulas

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0but i'm not introduced to them.....formulas!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i.e. the formulas to find the distance travelled as a function of time

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0to find the time only.....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes this is a multi step problem and my way of solving it was to find the time it falls.... multiply by the horizontal velocity and that gives you the horizontal distance travelled

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0so if i apply the way i do the time problems i should be able to get 5.52sec right?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0never mind i get 140 sec or 2.3 mintues....

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.05.52 seems less for some reason.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea well reason this out my acceleration is approximately 10 meter per second per second and we are only travcelling 150m

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0soooo if the fall time is 140 seconds.... clearly that would be wayyyyyyyyyyyyyyyyy more than 150m

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0ok..... can you please write out the equation you used to find the time..... PLease

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0look above.... i already listed them

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0so....x(t)=final velocity + initail velocit + 1/2(gxt^2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0did you get the equations i actaully worked out??? when i solved for T?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0i can solve the equation...no problem....but help me label this.... our initial velocity is 40m/sec, our final is 0m/sec?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and I found it to be T=5.53 seconds

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0are my initial and final ok????

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no... the INITIAL x velocity is 40 m/s and y is 0m/s... the FINAL x velocity is still 40 m/s and the y velocity you can solve for if you like but I didnt

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0and how far ahead of the campsite should the medical supplies be dropped?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441077238617:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so if you take that time value I solved for.... and simply multiply by the x velocity according to the equation given above: \[x(T)=v_{i_{x}}T=(40 \frac{m}{s})*5.53s=221m\]

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0ok Thank you very very much, and god bless you!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok wait wait yum i really want to know that this made sense... could you please explain to me what we just did plz :)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0ok so gosh....LOL ok :D

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0the only thing i didn't understand was wy idn't we use the following equation to solve for time....dv = vi X t + 1/2(a X t^2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because what does dv represent here? pretend like you are explaining how to solve the problem to me

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0dv is the vertical displacement from the ground up...150m at which the plane is flying at.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0again pretend like i dont understand how to solve this problem and walk me through it please :)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0ok we have the following variables already given to us, the initial velocity of the plane = 40m/sec, dv = 150, g = 9.81m/sec^2 and the final velocity = 0m/sec

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0actually the question didn't mention the final velocity so take that out. Now the only thing that is missing here is the time...and to solve for the time the only equations that i know of are the following..

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0give me a sec please......

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0yes ok ....i got it..wait

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0change in displacement = initial velocity X t + 1/2 a X t^2

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0this is probably the only equation i can think of to solve for the time....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes that is a nice list of formulas but explain to me what we did please

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0ok so we're going to find the time in which the package covered both vertical and horizontal distance....

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0but since we don't know the vertical distance from the plane to the ground, but since we have the initial velocity, gravitational force and vertical displacement...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you mean since we dont know the horizontal distance we cover while the package is falling.... other than that good keep going

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0we can figure out with help of the constant gravity how long it'll take the package to touch down to the ground.

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0so there is this plane flying forward with the speed of 40m/sec, then there is the gravitation force being applied on the package....9.81m/sec^2....

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0what do we do now....O_o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so what I was looking for was close

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we know that gravitational acceleration.... initial velocity in both the x and y directions.... and we define the initial positions to be zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also we know the altitude of the plane

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0oh i forgot to mention one more thing....let me finish

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so the logic is that while this thing is falling... no forces act on it in a horizontal direction... meaning the x velocity remains constant during the fall

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the y velocity however keeps increasing because of the force of gravity (i.e. the presence of a gravitational acceleration g)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0since we know our initial velocity and we know that the final will be zero m/sec, and we also know that the gravitational force is 9.81m/sec^2 is being applied on the object....we need to find what time it takes to get from that 40m/sec down to 0m/sec.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we know that with all the givens we can find the time it takes to fall (it is the only unknown in the equation listed above for y(t)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we then use that time to plug into the x equation so that this equation now only has one unknown

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yum where are you getting the final velocity from?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0Vertical: Horizontal: g = 9.8m/sec dh =? dv = 150m vh =? vi = 40m/sec t = ? vf = 0m/sec t = ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did you find that though?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0well i assume that once the package hit the ground, it's velocity will be 0m/sec

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0the package was dropped from the plane at 40m/sec. but it hits the ground, it'll be 0m/sec.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea in reality it goes to 0 m/s but what I am getting at is where in the formulas did the final velocity come in to play?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0oh that's the format i use when i solve these kinds of question.......it doesn't mean every single variable has to go into the formula.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea exactly.... writing down everything you know is very important but one thing that comes with experience is knowing what to use and what is irrelevant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now there are other ways to solve this problem that perhaps use that information, but this is the most straightforward way using the equations of motion

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0can you tell me one other way please....the easiest way :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is the easiest way yum... the only "easier" way is to algebraically solve one equation and subsitute it into the other.... but you are still using the same two equations and doing the same procedure.... only difference is you didnt solve for the actual time T in between

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0ok. Thanks and Thanks!!! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no problem yum just go over this logic again... set aside the answer and work it out... it may seem like a trivial exercise but its problems like these that allow you to make mistakes early on so later when problems are harder you dont struggle with them
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