- YumYum247

Help ME Please!!!!!!!! :)

- schrodinger

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- YumYum247

- YumYum247

Hi there please help me!!!

- anonymous

Hello give me one second..

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## More answers

- YumYum247

ok take your time!!! :)

- anonymous

ok lemme see

- 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

Potential force O_o

- anonymous

ok what else?

- anonymous

so the potential ENERGY the object has is given by what force? lets start there

- 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

yes yes exactley

- anonymous

so the FORCES on the object in the plane are ?

- YumYum247

because as the package is dropped from the plane, it'll travel both in the vertical and horizontal axis....

- anonymous

dont overthink it come on

- anonymous

yes yes but the forces?

- YumYum247

|dw:1441073471996:dw|

- YumYum247

potential and kinectic O_o

- 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

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

and as you said before we have two axes in which we are travelling vertical and horizontal

- YumYum247

Oh.....so that's what you meant so -9.81m/sec^2

- anonymous

so gravity is the only force in the vertical direction... yes exactley thats g the ACCELERATION due to gravity

- anonymous

what about the force in the horizontal direction? you already mentioned it was on a plane right?

- YumYum247

but there is the potential force of the plane dragging it on the horizontal axis to an angle.

- YumYum247

i'm referring to the package tho.

- YumYum247

so what's next LOL

- YumYum247

i think we first need to calculate how far on the horizontal axis the medical supplies are dropped?

- YumYum247

from the campsite.....

- anonymous

perhaps but I am trying to walk you through how you should think about these things in general

- YumYum247

|dw:1441073888474:dw|

- YumYum247

ok :)

- anonymous

Yes how you are thinking about it geometrically is valid dont take what I am saying as you are wrong

- 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

so first step is to consider the forces on the package

- anonymous

well for starters there is gravity acting down on the package through its center of mass

- YumYum247

is there a step wise procedure to get at these types of problems??????

- anonymous

but there is also the force of the plane's engines on the package

- YumYum247

true!!

- 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

yea :D that was what i was getting at

- anonymous

if the plane was applying a force to the package then the package wouldnt be flying along with the plane no?

- anonymous

however once the package is thrown out of the plane.... what forces act on it in a horizontal direction

- YumYum247

kinetic force.

- YumYum247

gravitational force!!!

- anonymous

no im sorry yum kinetic is not a force

- anonymous

yes gravity is still acting on it but that is in the vertical direction

- YumYum247

yeah i thought about that too late.....:D

- anonymous

you are confusing the two concepts of force and kinetic energy

- anonymous

its alright no worries :D

- YumYum247

:D sorry pal i'll try not to do it again!!

- 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

force i mean

- YumYum247

gravitational force is acting on the package.

- anonymous

correct.... and is there anything acting on the package horizontally?

- anonymous

btw no is a valid answer...

- YumYum247

would that be gravity as well. because the package is in the free fall, the only thing there is is gravity!!!

- YumYum247

i think the gravity is making the package distace both vertically and horizontally.

- anonymous

YES!!! EXACTLEY :D :D

- anonymous

no horizontal force

- anonymous

soooooo what does that mean for the horizontal veloctiy?

- 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

so the horizontal component is constant but the vertical component is changing.

- anonymous

YES AWESOME!!!

- 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

during the time in which it is falling

- YumYum247

yes yes!! :D

- YumYum247

so how do we figure out how far ahead of the campsite the medical packages were dropped????

- YumYum247

let me draw...

- anonymous

ok brilliant we know the distance between the campsite and the plane (vertical distance)

- anonymous

we know the initial VERTICAL velocity (pre fall) was zero (because the plane was neither climbing nor falling)

- YumYum247

|dw:1441074773233:dw|

- anonymous

and we know the ACCELERATION the package undergoes VERTICALLY while falling

- YumYum247

-9.8m/sec^2

- anonymous

Yes so how do we calculate the time of fall?

- 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

no in order to figure out the horizontal distance we first need to figure out the time of fall

- anonymous

in other words the time it takes to cover the vertical distance.... that we know

- anonymous

do you know how to set that equation up?

- YumYum247

let me try first....:)

- YumYum247

Change in vertical displacement = initial velocity X t + 1/2(g X t^2) and solve for time....

- YumYum247

do you want me to draw for you the equation??????????

- YumYum247

|dw:1441075219970:dw|

- anonymous

|dw:1441075283811:dw|

- anonymous

so here we will take the intial position to be zero

- YumYum247

|dw:1441075275924:dw|

- anonymous

no no dont do that....

- YumYum247

whyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy!!!?!?!?!?!

- anonymous

you are confusing the initial velocity int he x direction for an initial velocity in the y direction....hold on:

- YumYum247

ok so i have to figure out the vertical component and horizontal component???

- YumYum247

i mean do i have to.....???

- YumYum247

but need the angle for that :(

- 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

ok so we know y(T)=150m (the total fall height)

- YumYum247

ok.

- 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

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

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

its a good exercise in unit analysis to check for yourself

- anonymous

yes very good idea at the begining of every problem.... write down your knowns and your unknowns

- YumYum247

but idk why you applied that formula tho :O

- YumYum247

wait let me try doing it my wait and see if i get 5.52sec.
Hold up, din't go anywhere!!! :)

- anonymous

which formula? those are the kinetic motion formulas

- YumYum247

but i'm not introduced to them.....formulas!!!

- anonymous

i.e. the formulas to find the distance travelled as a function of time

- YumYum247

to find the time only.....

- 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

so if i apply the way i do the time problems i should be able to get 5.52sec right?

- YumYum247

give me a sec.....

- YumYum247

never mind i get 140 sec or 2.3 mintues....

- YumYum247

5.52 seems less for some reason.

- anonymous

yea well reason this out my acceleration is approximately 10 meter per second per second and we are only travcelling 150m

- anonymous

soooo if the fall time is 140 seconds.... clearly that would be wayyyyyyyyyyyyyyyyy more than 150m

- YumYum247

ok..... can you please write out the equation you used to find the time..... PLease

- anonymous

look above.... i already listed them

- YumYum247

so....x(t)=final velocity + initail velocit + 1/2(gxt^2)

- anonymous

did you get the equations i actaully worked out??? when i solved for T?

- 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

and I found it to be T=5.53 seconds

- YumYum247

are my initial and final ok????

- YumYum247

Vi and vf?

- 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

ok...

- YumYum247

and how far ahead of the campsite should the medical supplies be dropped?

- YumYum247

|dw:1441077238617:dw|

- 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

ok Thank you very very much, and god bless you!!!

- 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

ok so gosh....LOL ok :D

- 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

because what does dv represent here? pretend like you are explaining how to solve the problem to me

- YumYum247

dv is the vertical displacement from the ground up...150m at which the plane is flying at.

- anonymous

again pretend like i dont understand how to solve this problem and walk me through it please :)

- 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

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

give me a sec please......

- YumYum247

yes ok ....i got it..wait

- YumYum247

change in displacement = initial velocity X t + 1/2 a X t^2

- YumYum247

this is probably the only equation i can think of to solve for the time....

- YumYum247

##### 1 Attachment

- anonymous

yes that is a nice list of formulas but explain to me what we did please

- YumYum247

ok so we're going to find the time in which the package covered both vertical and horizontal distance....

- anonymous

ok

- 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

you mean since we dont know the horizontal distance we cover while the package is falling.... other than that good keep going

- 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

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

ok keep doing

- YumYum247

what do we do now....O_o

- anonymous

ok so what I was looking for was close

- 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

also we know the altitude of the plane

- YumYum247

oh i forgot to mention one more thing....let me finish

- 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

the y velocity however keeps increasing because of the force of gravity (i.e. the presence of a gravitational acceleration g)

- 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

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

we then use that time to plug into the x equation so that this equation now only has one unknown

- anonymous

yum where are you getting the final velocity from?

- YumYum247

Vertical: Horizontal:
g = -9.8m/sec dh =?
dv = 150m vh =?
vi = 40m/sec t = ?
vf = 0m/sec
t = ?

- anonymous

how did you find that though?

- YumYum247

well i assume that once the package hit the ground, it's velocity will be 0m/sec

- YumYum247

the package was dropped from the plane at 40m/sec. but it hits the ground, it'll be 0m/sec.

- 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

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

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

yah your right :(

- 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

ok :D

- YumYum247

can you tell me one other way please....the easiest way :(

- 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

ok. Thanks and Thanks!!! :)

- 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

yes i agree!!! :)

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