Bomber
A moving jet fighter has released a bomb.
What is the path of the bomb as observed by a stationary solider on the ground?
Neglect air resistance.
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5 solutions
But what if the soldier is standing in the path of the bomber plane...
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It is still a parabola with x = 0 .
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"a parabola with
x
=
0
" - Chew-Seong Cheong
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@Jonathan Quarrie – Yes. Precisely.
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@Chew-Seong Cheong – Are asserting that there is no such thing as a straight line?
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@Jonathan Quarrie – There is such a thing as a straight line. But a straight line can be a parabola, a circle, a curve, because a two-dimensional object with one-dimension suppressed becomes a straight line. Just line a circle in two dimensions can be a sphere, a cylinder, a ellipsoid, ... in three dimensions with one dimension suppressed.
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@Chew-Seong Cheong – Would you agree that there is a distinction between a parabola and a parabola with one-dimension suppressed?
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@Jonathan Quarrie – Of course there is distinction between a parabola and a straight line. But in this case it is a parabola. The trajectory of the bomb is the same only the observer is at a special position. When the observer is an infinitesimal away from x = 0 it is a parabola. And how small is infinitesimal?
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@Chew-Seong Cheong – Indeed the trajectory of the bomb remains the same, but the problem does not ask us about the trajectory. It asks us to describe the observed path.
Are you saying that an observer can never be at x = 0 ?
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@Jonathan Quarrie – Of course, it can. But I was answering for the general case of y = sin 2 θ a x 2 , where θ is the angle between the path of the airplane and the observer. The special case you mentioned is when θ → 0 . But the relation y = sin 2 θ a x 2 is still parabolic. If we use parametric function of y = 2 1 g t 2 , there is a difference with a straight line of x = 0 where y is not clearly defined.
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@Chew-Seong Cheong – I would go as far to say that - by not eliminating special cases - it is this problem that's not clearly defined.
But it also seems like your position is that you are concerned with the process of drawing a straight line (parabolic/linear), rather than the result being a straight line.
If I were to take a turn on that position, I should not answer "
0
" when asked what
1
−
1
equals - I should answer "
1
−
1
", because it describes the process.
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@Jonathan Quarrie – But it is clear to me. General solutions exist in many cases including this one. Solution should not be affected by frame of reference like Einstein said.
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@Chew-Seong Cheong – Had the question been purely about the trajectory of the bomb, I would agree. But the introduction of an observer necessitates a frame of reference.
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@Jonathan Quarrie – I think we have spent enough time on this. To me it is very trivial and the special case is unimportant. Why don't you spend more time studying.
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@Chew-Seong Cheong – While I admit that how I introduced myself into this conversation could have been done with a touch more class, I don't think your last sentence was necessary.
As a prominent figure on this site, you know as well as anyone that a single case within the constraints of a problem that contradicts an assertion will render the assertion false, no matter how trivial that case may be.
The wording of this problem has not eliminated this special case from possibility.
I appreciate the mental exercise in debating different opinions, as another form of self-improvement.
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@Jonathan Quarrie – Sorry, I mistook you for another person.
They wouldn't see anything. They would be dead.
They would either die of the fall or die of being hit by the bomber
Why would he do that? It ll be similar to suicide.
You are actually correct. The question is poorly worded. If the viewer is perpendicular to the path of the bomb he sees something approximating a parabola. But the rotation of the earth in relation to the bomb’s path will also affect that shape. If the observer is standing directly in like with the falling bomb, the observer will see a straight line. The observer doesn’t have to die. It may be standing miles aft of the plane (in the direction from which the plane came) so the bomb will land no where near the observer.
I am not a science student. That's why i am asking ∫ x d t = ( 2 ∗ v h ) ( v h 2 ∗ t 2 ) But your answer is 2 ∗ v h 2 ( g ∗ v h 2 ∗ t 2 ) . How is that?
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It was not ∫ x d t . That is why. y = 2 1 g t 2 and x = v h t ⟹ t = v h x , then y = 2 v h 2 g x 2 .
Your observation about crewman is spot on and it would have been more interesting had it that been the question instead. To add colour to it you could have made the plane go slower or faster after dropping the bomb.
Basically it has a horizontal and vertical motion, outlining projectile motion which is parabolic motion.
Amazing explanations!
I have sent this question to my relatives in Syria, they will get back to me with first hand eye witness experience of this on the receiving end. I pray that they survive if it so that you may all have a definitive answer.
But it's not a porabola. Nobody even noticed. It's an exponential function -f(-x)
It will be parabolic as after every second increment its horizontal distance will be same but increment in vertical distance will not same ( will increases)after every second as it is accelerating at 9.8m/s² due to gravity. For ex-
• Let the speed of jet be 1 m/s and at a height of 45 m
•Let this scene starts from origin and happen in 3rd quadrant of graph.
• Then after one second coordinate will be A(-1,-1/2 gt²=5) or A(-1,-5)
•Similarly,after 2 second the coordinate will be B(-2,-20) And ,after 3 second the coordinate will be C(-3,-45)
•Joining this coordinate will give us parabolic path.
If presenting a object in a graph maybe we could change some measurements also based same as the graph. Like for example The height of the Plane is ( 45m ), we could just turn it into a location of the grid instesd of using a unit not existing in the graph. But i like your concept its an easy explanation.
It will be a projectile motion
It would be parabola. I support Ram.
lol the title is so silly XD
You should clarify where the observer is relative to the plane. I got it right but if the observer were in front of the plane he would see a straight line. If a side view he sees a parabola. Nobody wants to guess what the problem writer intended. As soon as you provide the straight line option the question is where is the observer.
The Main concept used in this Question is INERTIA. It is the property of a body to remain in the state of Motion or in the state of Rest unless any External Force is applied to the object.
Now, as the Plane was moving forward, It was in Motion and in a particular Direction (Forward), then after the Bomb is just dropped, It will also try to be in the state of Motion in the same direction (Forward) as of Plane, due to Both Inertia of Motion and Inertia of Direction. But, as the force of Gravity acts Downwards, so It will possess both the Horizontal Component i.e. while moving forward within the state of Motion, and the Vertical Component i.e. while moving downwards due to Acceleration due to Gravity (g=9.8m/s^2), making It follow a Parabolic Path or Projectile Motion in the Forward Direction.
hell ms math dead eye still
It will have vertical motion due to the gravity and horizontal motion due to the force given by the air thus resulting in a projectile motion. The path covered will be a parabola.
When the bomb is released it has the velocity of the plane v h , which is horizontal. Neglecting air resistance, the bomb continues to move horizontally at v h according to Newton's first law and the horizontal displacement after time t is x = v h t . Assuming the plane is flying at a constant velocity, the bomb is at all times on a same vertical line underneath the plane (see the figure). A crewman in the plane would notice that the bomb is falling vertically down.
Once the bomb is released it is subject to the acceleration due to gravity ≈ 9 . 8 ms − 2 . Its vertical displacement from the plane after time t is y = 2 1 g t 2 = 2 v h 2 g x 2 , which is a parabola .
Reference: Check out this YouTube video .