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Imagine an aeroplane travelling with velocity $v$ at some angle $\alpha$ from East to North. A box is dropped from the aeroplane.

What would the projectile of the box be? Would it be a parabola with an initial y-component of velocity acting upwards, or would it travel as a downward curve (with an initial y-component of velocity acting downwards)?

My reasons for thinking both:

Parabolic projection

My explanation for this would be that the box will have the same initial components of velocity as the aeroplane at the time it is dropped. Since the aeroplane has a y-component of velocity acting upwards and an x-component of velocity it would travel in a parabola, as in conventional projectile motion.

Downward curve

(By "downward curve" I mean a curve with negative gradient increasing in magnitude.) My explanation for this would be that as soon at it is let go the only force acting on the box would be its weight, and hence it couldn't travel upwards. As well as this, when I visualise in my head a box being dropped from an aeroplane, I can't imagine it's possible to travel in a diagonally upward direction once it's been let go.

Can someone explain to me which type of motion it has and why?

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  • $\begingroup$ Note that the angle is redundant, the velocity already specifies the angle. $\endgroup$
    – user541686
    Commented Oct 7, 2014 at 11:07

5 Answers 5

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It will travel along a parabola (ignoring drag from the air here), initially with upward velocity, as you describe in your first scenario.

You're correct that the only force acting on the box is its weight, but this means it will have downward acceleration immediately, not necessarily downward velocity. Eventually the downward acceleration will lead to negative velocity, but initially the box travels with the same velocity as the plane.

Travelling upward once it's been let go isn't strange at all. If you throw a ball into the air, it travels upward for a while before beginning to drop, and this isn't strange...

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  • $\begingroup$ Brilliant to use the ball analogy -- your hand is initially travelling upward, and when you "drop" the ball, it continues upward until the downward acceleration overcomes the upward velocity. Simple, clear example that even a child can understand :) $\endgroup$
    – Doktor J
    Commented Oct 7, 2014 at 16:05
  • $\begingroup$ Wouldn't it be better if you prove this mathematically? $\endgroup$
    – soumyadeep
    Commented Oct 7, 2014 at 19:05
  • $\begingroup$ @soumyadeep Better? Perhaps by some measure of 'better'. My assessment of the OP's question was that a mathematical explanation might not resolve the confusion, while an intuitive one might be 'better'. Besides, if you want the math, virtually any google hit for 'projectile motion' will be happy to oblige. I'd argue that intuition is harder to come by. $\endgroup$
    – Kyle Oman
    Commented Oct 7, 2014 at 21:40
  • $\begingroup$ @Kyle I just said that that because i think we cannot prove it will be a parabola or not any other curve.But whatever, as you said,the focus of OP's question is why it will go up-I did not notice that earlier-sorry. $\endgroup$
    – soumyadeep
    Commented Oct 8, 2014 at 2:45
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Your question has great practical significance: it is the very essense of loft bombing/LABS.

Since nuclear weapons can damage/destroy an attacking plane, it was deemed necessary to devise a method to release the bomb and to increase the separation between the air burst and the aircraft.

I recommend watching this training video: http://www.youtube.com/watch?v=3dIqfN_aPtY

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    $\begingroup$ What an excellent video to describe the answer, and disturbing on quite a few levels as well. $\endgroup$
    – crthompson
    Commented Oct 7, 2014 at 17:08
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The accepted answer is (subtly) wrong. While the projectile will indeed initially have an upward velocity, the shape it traces out is that of an ellipse, not a parabola -- remember, the earth is not flat.

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    $\begingroup$ This is true, but then I do mention the caveat of ignoring drag from the air, which is almost certainly a bigger caveat than accounting for the curvature of the Earth. $\endgroup$
    – Kyle Oman
    Commented Oct 7, 2014 at 17:38
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    $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. $\endgroup$
    – ACuriousMind
    Commented Oct 7, 2014 at 17:55
  • $\begingroup$ @ACuriousMind: How is this not an answer? The question is what the shape would be and I clearly say that it's an ellipse... $\endgroup$
    – user541686
    Commented Oct 7, 2014 at 18:33
  • $\begingroup$ @Kyle: I mentioned it because it's a common misconception that I felt people on this page had fallen for and was worth clearing up, not because I thought it was "more of a caveat" than air drag in some way. $\endgroup$
    – user541686
    Commented Oct 7, 2014 at 18:36
  • $\begingroup$ It's not a "misconception" it is a consequence of using the constant field approximation for small motions near the surface of the Earth. And it is a very good approximation with multiple things going for it chief among them being that the math is tractable for beginners. $\endgroup$ Commented Oct 8, 2014 at 3:23
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All objects that are in free fall (which is to say, no other forces are acting on said objects) will have a height above ground that can be predicted as such, until said objects reach the ground:

h=-(g/2)(t^2) + vt + c

where: g represents the acceleration due to gravity. On Earth, at sea level, this value is approximately 9.80665 m/s^2. t represents the time elapsed since the fall of the object(s) in question began. v represents the initial vertical velocity. c represents the initial height, usually above ground, of the object(s) in question.

Therefore, the answer is (a): parabola.

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You're missing perspective in your question.

With respect to the Earth, the box will travel in a parabolic arc, just as if you've thrown the box into the air with the same initial velocity. While the acceleration is certainly downwards after you drop it, its velocity is still upwards (and forward) until gravity changes that (and gravity is quite weak, so it takes a lot more time than e.g. bouncing the box against a wall).

With respect to the plane, it will drop straight down (and backwards, thanks to air drag), as simple as that.

Which of these is the right one? Just go back to Newton and Galileo - those two perspectives are simply two different inertial frames of reference. They're both completely consistent ways of looking at the world, and they both fit just fine - in fact, you're only doing a linear transformation - when looking in the plane's reference frame, you've simply subtracted the velocity of the plane from the velocity of the box, both with respect to the Earth.

So the answer is: It depends on your frame of reference. But both are, and have to be, consistent (interesting implications of that can be found in the special theory of relativity).

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