YumYum247
  • YumYum247
Help ME Please!!!!!!!! :)
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
YumYum247
  • YumYum247
@PlasmaFuzer
YumYum247
  • YumYum247
Hi there please help me!!!
anonymous
  • anonymous
Hello give me one second..

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YumYum247
  • YumYum247
ok take your time!!! :)
anonymous
  • anonymous
ok lemme see
anonymous
  • anonymous
Well 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?
YumYum247
  • YumYum247
Potential force O_o
anonymous
  • anonymous
ok what else?
anonymous
  • anonymous
so the potential ENERGY the object has is given by what force? lets start there
YumYum247
  • YumYum247
the 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
  • anonymous
yes yes exactley
anonymous
  • anonymous
so the FORCES on the object in the plane are ?
YumYum247
  • YumYum247
because as the package is dropped from the plane, it'll travel both in the vertical and horizontal axis....
anonymous
  • anonymous
dont overthink it come on
anonymous
  • anonymous
yes yes but the forces?
YumYum247
  • YumYum247
|dw:1441073471996:dw|
YumYum247
  • YumYum247
potential and kinectic O_o
anonymous
  • anonymous
those are the energys.... ok one of the FORCES is gravity right? gravity is always acting on everythin.... unless we flying in space right?
YumYum247
  • YumYum247
when 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
  • anonymous
and as you said before we have two axes in which we are travelling vertical and horizontal
YumYum247
  • YumYum247
Oh.....so that's what you meant so -9.81m/sec^2
anonymous
  • anonymous
so gravity is the only force in the vertical direction... yes exactley thats g the ACCELERATION due to gravity
anonymous
  • anonymous
what about the force in the horizontal direction? you already mentioned it was on a plane right?
YumYum247
  • YumYum247
but there is the potential force of the plane dragging it on the horizontal axis to an angle.
YumYum247
  • YumYum247
i'm referring to the package tho.
YumYum247
  • YumYum247
so what's next LOL
YumYum247
  • YumYum247
i think we first need to calculate how far on the horizontal axis the medical supplies are dropped?
YumYum247
  • YumYum247
from the campsite.....
anonymous
  • anonymous
perhaps but I am trying to walk you through how you should think about these things in general
YumYum247
  • YumYum247
|dw:1441073888474:dw|
YumYum247
  • YumYum247
ok :)
anonymous
  • anonymous
Yes how you are thinking about it geometrically is valid dont take what I am saying as you are wrong
anonymous
  • anonymous
i 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
  • anonymous
so first step is to consider the forces on the package
anonymous
  • anonymous
well for starters there is gravity acting down on the package through its center of mass
YumYum247
  • YumYum247
is there a step wise procedure to get at these types of problems??????
anonymous
  • anonymous
but there is also the force of the plane's engines on the package
YumYum247
  • YumYum247
true!!
anonymous
  • anonymous
i 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
  • anonymous
yea :D that was what i was getting at
anonymous
  • anonymous
if the plane was applying a force to the package then the package wouldnt be flying along with the plane no?
anonymous
  • anonymous
however once the package is thrown out of the plane.... what forces act on it in a horizontal direction
YumYum247
  • YumYum247
kinetic force.
YumYum247
  • YumYum247
gravitational force!!!
anonymous
  • anonymous
no im sorry yum kinetic is not a force
anonymous
  • anonymous
yes gravity is still acting on it but that is in the vertical direction
YumYum247
  • YumYum247
yeah i thought about that too late.....:D
anonymous
  • anonymous
you are confusing the two concepts of force and kinetic energy
anonymous
  • anonymous
its alright no worries :D
YumYum247
  • YumYum247
:D sorry pal i'll try not to do it again!!
anonymous
  • anonymous
dont worry about it... ok so think really hard what is acting on the package HORIZONTALLY once its been thrown out of the plane
anonymous
  • anonymous
force i mean
YumYum247
  • YumYum247
gravitational force is acting on the package.
anonymous
  • anonymous
correct.... and is there anything acting on the package horizontally?
anonymous
  • anonymous
btw no is a valid answer...
YumYum247
  • YumYum247
would that be gravity as well. because the package is in the free fall, the only thing there is is gravity!!!
YumYum247
  • YumYum247
i think the gravity is making the package distace both vertically and horizontally.
anonymous
  • anonymous
YES!!! EXACTLEY :D :D
anonymous
  • anonymous
no horizontal force
anonymous
  • anonymous
soooooo what does that mean for the horizontal veloctiy?
YumYum247
  • YumYum247
the 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
  • YumYum247
so the horizontal component is constant but the vertical component is changing.
anonymous
  • anonymous
YES AWESOME!!!
anonymous
  • anonymous
ok 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
  • anonymous
during the time in which it is falling
YumYum247
  • YumYum247
yes yes!! :D
YumYum247
  • YumYum247
so how do we figure out how far ahead of the campsite the medical packages were dropped????
YumYum247
  • YumYum247
let me draw...
anonymous
  • anonymous
ok brilliant we know the distance between the campsite and the plane (vertical distance)
anonymous
  • anonymous
we know the initial VERTICAL velocity (pre fall) was zero (because the plane was neither climbing nor falling)
YumYum247
  • YumYum247
|dw:1441074773233:dw|
anonymous
  • anonymous
and we know the ACCELERATION the package undergoes VERTICALLY while falling
YumYum247
  • YumYum247
-9.8m/sec^2
anonymous
  • anonymous
Yes so how do we calculate the time of fall?
YumYum247
  • YumYum247
ok 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
  • anonymous
no in order to figure out the horizontal distance we first need to figure out the time of fall
anonymous
  • anonymous
in other words the time it takes to cover the vertical distance.... that we know
anonymous
  • anonymous
do you know how to set that equation up?
YumYum247
  • YumYum247
let me try first....:)
YumYum247
  • YumYum247
Change in vertical displacement = initial velocity X t + 1/2(g X t^2) and solve for time....
YumYum247
  • YumYum247
do you want me to draw for you the equation??????????
YumYum247
  • YumYum247
|dw:1441075219970:dw|
anonymous
  • anonymous
|dw:1441075283811:dw|
anonymous
  • anonymous
so here we will take the intial position to be zero
YumYum247
  • YumYum247
|dw:1441075275924:dw|
anonymous
  • anonymous
no no dont do that....
YumYum247
  • YumYum247
whyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy!!!?!?!?!?!
anonymous
  • anonymous
you are confusing the initial velocity int he x direction for an initial velocity in the y direction....hold on:
YumYum247
  • YumYum247
ok so i have to figure out the vertical component and horizontal component???
YumYum247
  • YumYum247
i mean do i have to.....???
YumYum247
  • YumYum247
but need the angle for that :(
anonymous
  • anonymous
\[x(t)=x_{0}+v_{i_{x}} t + \frac{1}{2}a t^{2} = v_{i_{x}}\] && \[y(t)=y_{0}+v_{i_{y}} t + \frac{1}{2}a t^{2} =-\frac{1}{2}g t^{2} \]
anonymous
  • anonymous
ok so we know y(T)=150m (the total fall height)
YumYum247
  • YumYum247
ok.
anonymous
  • anonymous
so 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
  • anonymous
How 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
  • YumYum247
wait 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
  • anonymous
its a good exercise in unit analysis to check for yourself
anonymous
  • anonymous
yes very good idea at the begining of every problem.... write down your knowns and your unknowns
YumYum247
  • YumYum247
but idk why you applied that formula tho :O
YumYum247
  • YumYum247
wait let me try doing it my wait and see if i get 5.52sec. Hold up, din't go anywhere!!! :)
anonymous
  • anonymous
which formula? those are the kinetic motion formulas
YumYum247
  • YumYum247
but i'm not introduced to them.....formulas!!!
anonymous
  • anonymous
i.e. the formulas to find the distance travelled as a function of time
YumYum247
  • YumYum247
to find the time only.....
anonymous
  • anonymous
yes 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
  • YumYum247
so if i apply the way i do the time problems i should be able to get 5.52sec right?
YumYum247
  • YumYum247
give me a sec.....
YumYum247
  • YumYum247
never mind i get 140 sec or 2.3 mintues....
YumYum247
  • YumYum247
5.52 seems less for some reason.
anonymous
  • anonymous
yea well reason this out my acceleration is approximately 10 meter per second per second and we are only travcelling 150m
anonymous
  • anonymous
soooo if the fall time is 140 seconds.... clearly that would be wayyyyyyyyyyyyyyyyy more than 150m
YumYum247
  • YumYum247
ok..... can you please write out the equation you used to find the time..... PLease
anonymous
  • anonymous
look above.... i already listed them
YumYum247
  • YumYum247
so....x(t)=final velocity + initail velocit + 1/2(gxt^2)
anonymous
  • anonymous
did you get the equations i actaully worked out??? when i solved for T?
YumYum247
  • YumYum247
i can solve the equation...no problem....but help me label this.... our initial velocity is 40m/sec, our final is 0m/sec?
anonymous
  • anonymous
and I found it to be T=5.53 seconds
YumYum247
  • YumYum247
are my initial and final ok????
YumYum247
  • YumYum247
Vi and vf?
anonymous
  • anonymous
no... 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
  • YumYum247
ok...
YumYum247
  • YumYum247
and how far ahead of the campsite should the medical supplies be dropped?
YumYum247
  • YumYum247
|dw:1441077238617:dw|
anonymous
  • anonymous
so 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
  • YumYum247
ok Thank you very very much, and god bless you!!!
anonymous
  • anonymous
ok wait wait yum i really want to know that this made sense... could you please explain to me what we just did plz :)
YumYum247
  • YumYum247
ok so gosh....LOL ok :D
YumYum247
  • YumYum247
the 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
  • anonymous
because what does dv represent here? pretend like you are explaining how to solve the problem to me
YumYum247
  • YumYum247
dv is the vertical displacement from the ground up...150m at which the plane is flying at.
anonymous
  • anonymous
again pretend like i dont understand how to solve this problem and walk me through it please :)
YumYum247
  • YumYum247
ok 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
  • YumYum247
actually 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
  • YumYum247
give me a sec please......
YumYum247
  • YumYum247
yes ok ....i got it..wait
YumYum247
  • YumYum247
change in displacement = initial velocity X t + 1/2 a X t^2
YumYum247
  • YumYum247
this is probably the only equation i can think of to solve for the time....
YumYum247
  • YumYum247
anonymous
  • anonymous
yes that is a nice list of formulas but explain to me what we did please
YumYum247
  • YumYum247
ok so we're going to find the time in which the package covered both vertical and horizontal distance....
anonymous
  • anonymous
ok
YumYum247
  • YumYum247
but 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
  • anonymous
you mean since we dont know the horizontal distance we cover while the package is falling.... other than that good keep going
YumYum247
  • YumYum247
we can figure out with help of the constant gravity how long it'll take the package to touch down to the ground.
YumYum247
  • YumYum247
so 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....
anonymous
  • anonymous
ok keep doing
YumYum247
  • YumYum247
what do we do now....O_o
anonymous
  • anonymous
ok so what I was looking for was close
anonymous
  • anonymous
we know that gravitational acceleration.... initial velocity in both the x and y directions.... and we define the initial positions to be zero
anonymous
  • anonymous
also we know the altitude of the plane
YumYum247
  • YumYum247
oh i forgot to mention one more thing....let me finish
anonymous
  • anonymous
ok 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
  • anonymous
the y velocity however keeps increasing because of the force of gravity (i.e. the presence of a gravitational acceleration g)
YumYum247
  • YumYum247
since 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
  • anonymous
so 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
  • anonymous
we then use that time to plug into the x equation so that this equation now only has one unknown
anonymous
  • anonymous
yum where are you getting the final velocity from?
YumYum247
  • YumYum247
Vertical: Horizontal: g = -9.8m/sec dh =? dv = 150m vh =? vi = 40m/sec t = ? vf = 0m/sec t = ?
anonymous
  • anonymous
how did you find that though?
YumYum247
  • YumYum247
well i assume that once the package hit the ground, it's velocity will be 0m/sec
YumYum247
  • YumYum247
the package was dropped from the plane at 40m/sec. but it hits the ground, it'll be 0m/sec.
anonymous
  • anonymous
yea 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
  • YumYum247
oh 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
  • anonymous
yea 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
YumYum247
  • YumYum247
yah your right :(
anonymous
  • anonymous
now 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
  • YumYum247
ok :D
YumYum247
  • YumYum247
can you tell me one other way please....the easiest way :(
anonymous
  • anonymous
that 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
  • YumYum247
ok. Thanks and Thanks!!! :)
anonymous
  • anonymous
no 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
YumYum247
  • YumYum247
yes i agree!!! :)

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