Projectile Motion, package has to land on a moving ship after being droped from plane

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  • #1
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Homework Statement


A plane flies at 900m and has to deliver a package. It flies at 180 m/s North at 15° below the horizontal. The ships velocity is 40 m/s and it is travelling due north. At what horizontal distance from the ship must the package be dropped for it to land on the ship?


Homework Equations


I know to use the equations of motion but I don't understand the problem.


The Attempt at a Solution


Is the height of the plane changing?
The ship and the plane only have a velocity in one direction, where as the package once released has velocity in two (x and y) right?
How do I even begin to set this problem up?
 

Answers and Replies

  • #2
lewando
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Is the height of the plane changing?
Yes, it is descending.
The ship and the plane only have a velocity in one direction, where as the package once released has velocity in two (x and y) right?
What is your coordinate system (what is x?, what is y?)? The plane, package, and ship are all travelling along the N/S axis, but the plane and the package are also travelling along the up/down axis.
How do I even begin to set this problem up?
Now that you understand the problem, what equations seem more relevant? Draw a diagram of what is going on.
 
  • #3
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The height of the plane is changing.

It's speed is of 180 m/s at 15 degrees below the horizontal. It is therefore losing altitude.

This is a 2d problem: you could set the x axis pointing to the north, and the y axis being the height.

Draw a diagram: it will drastically help you understand the problem
 
  • #4
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@lewando, the x axis would be the distance's between the nose of the plane and the rear of the ship. The y axis would be the difference in altitudes.

Okay, I drew a diagram
PRe4S.png


the velocity of the plane is given but it has an x and v component.
Vplane = 180 m/s

Thus the intial velocity of the package is:
X: Vplane*cosθ
Y: Vplane*sinθ + gt (because gravity is also accelerating the package).

Vship = 40m/s

Initially I thought of using the altitude and the y component of the package's velocity to determine the time. Then I was going to use this time to solve for the distance required in the x component.

Although, now that I know the altitude is changing, I have no idea on how to approach the problem.
 
  • #5
lewando
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Vplane = 180 m/s

Thus the intial velocity of the package is:
X: Vplane*cosθ
Y: Vplane*sinθ + gt (because gravity is also accelerating the package).
The "+gt" does not belong in the initial velocity expression.

Initially I thought of using the altitude and the y component of the package's velocity to determine the time. Then I was going to use this time to solve for the distance required in the x component.
Why don't you try that?

Although, now that I know the altitude is changing, I have no idea on how to approach the problem.
The altitude of the package is at the mercy of the constant acceleration due to gravity. You must have done similar free-fall problems, yes?
 
  • #6
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The "+gt" does not belong in the initial velocity expression.

Why don't you try that?

The altitude of the package is at the mercy of the constant acceleration due to gravity. You must have done similar free-fall problems, yes?

I see what you are saying, in the initial velocity it gt is not involved, but AFTER it is dropped, yes.

I think I have worked out a solution.
Bare with me.

1. Vpack (y component) = Vplane*sinθ + gt
2. ((Vpack y)^2 - (Vpack nought y)^2)/(2g) = dy (altitude from which the package has to fall) v pack y = 0, this is the point when the package is on the ship.
3. dy = vpack nought y * t +.5gt^2, use quadratic formula to solve for the time of the drop

4. d ship travels = v ship * tdrop
5. d drop x = v package x * tdrop

now I have no clue how to relate the two distances..., would the horizontal distance, just be (5.-4.)?

Also, my professor already has a PhD and is now going for another graduate degree so he has very little time to help us at all. Also, this is probably the 2nd or 3rd projectile motion problem he has assigned us :D

Gotta love college :)
 
  • #7
lewando
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I can see some of you confusion based on how the question is worded.
A plane flies at 900m and has to deliver a package. It flies at 180 m/s North at 15° below the horizontal. The ships velocity is 40 m/s and it is travelling due north. At what horizontal distance from the ship must the package be dropped for it to land on the ship?
I am assuming at launch point that the height of the package is 900m and that the plane is approaching the boat at the moment of release.
 
  • #8
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I can see some of you confusion based on how the question is worded.
I am assuming at launch point that the height of the package is 900m and that the plane is approaching the boat at the moment of release.


i think so, but originally i had thought the plane is constantly at an altitude of 900m
 
  • #9
lewando
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The plane is definitely descending and therefore a non-zero y-component of initial velocity as you have correctly determined.
A plane flies at 900m and has to deliver a package.
With minimal poetic license, can this not be interpreted as "a flying plane has to deliver a package at 900m"?

I think you have to fix the drop height otherwise you can't determine a horizontal distance. For example, if the plane's altitude was zero when the package was released then so would be the horizontal distance, and how cool would that be!
 
  • #10
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The plane is definitely descending and therefore a non-zero y-component of initial velocity as you have correctly determined.

With minimal poetic license, can this not be interpreted as "a flying plane has to deliver a package at 900m"?

I think you have to fix the drop height otherwise you can't determine a horizontal distance. For example, if the plane's altitude was zero when the package was released then so would be the horizontal distance, and how cool would that be!

That's what I had thought as well, that the height of the plane was fixed.
If i go with this approach, then the plane doesn't have two velocity components? Only one because the height is unchanging. Which would mean the package's velocity in the y component is due to gravity; it would be in free fall with velocity only in the x direction caused by the plane

However, I did use the quadratic formula to determine a time that the box would take to fall to altitude of 0. I like using a fixed altitude, much easier...oh well not my fault professor is ambiguous.
 
  • #11
lewando
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That's what I had thought as well, that the height of the plane was fixed.
What I meant was fixed (known, specified) at the time of the drop.

If i go with this approach, then the plane doesn't have two velocity components? Only one because the height is unchanging.
No--2 components. The plane's altitude is changing as it descends.
 
  • #12
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What I meant was fixed (known, specified) at the time of the drop.

No--2 components. The plane's altitude is changing as it descends.

That's it, I'm putting down what I originally had as my answer:

"Plane's do not drop packages. If this plane dropped the package, then the package would be destroyed."

But, we could assume the package is dropped from an altitude of 900m?
 
Last edited:
  • #13
lewando
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900m is a good assumption.
 
  • #14
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Why would you do that?

lol now I look like the crazy one
 
  • #15
lewando
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Well pick whatever the problem really states and go with that. Don't worry about your anonymity--this is 2011.
 

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