# Question that neither my teacher nor tutor could do (Grade 12 Physics)

• s3a

## Homework Statement

http://i.imgur.com/7DzVr.png

## Homework Equations

Kinematics Equations:
Vf = Vi + at
Δy = Vi(t) + 1/2 * a * t^2
Vf^2 = Vi^2 + 2aΔy
Δy = (Vi + Vf)/2 * t

## The Attempt at a Solution

My teacher and tutor don't know how to do this and some strangers told me that I had to measure the distance between each bomb with a ruler and I measured the length of the plane as well and the problem gives us the length of the plane so we can find the scale like that. My ruler gave length = 10.85mm and 1.5mm, 2.1mm, and 1.4mm respectively (the first one being the one farthest from the plane). I didn't count the rest of the bombs because it is irritating and I'm not even sure if I have to do that. If I really do then I'll measure the rest but this question counts for (few) marks even though my teacher doesn't know how to do it!

Any help would be GREATLY appreciated!

The question asks for the average distance between craters so you only need the average distance between the bombs. Divide total distance by 12. (or 11 if you start and end on a bomb!)
Do you know how to continue after that?

You do not need that many formulas to answer this question. First find a relationship between the fall time and vertical distance (one of the formulas you gave) for a bomb and use this on the first and last bomb to find the time difference between their release from their measured release distances, and then think about how this time difference relates to distance in their ground impact point. You may need to make some assumptions along the way.

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Thanks to both for your response. Before I continue further; Is Δy = 2000/11 = 181.82m/s ?

Thanks to both for your response. Before I continue further; Is Δy = 2000/11 = 181.82m/s ?

No. In your original post delta-y is a distance, but here you mention a speed? You also seem to involve the altitude of the plane (2000 m) which is not really needed.

Using mentioned formula for delta-y applied to the vertical distance from release point as seen from the plane, think about what the initial vertical speed is in that case. You should get a simple relationship between the height of a single bomb from its release height (which you can measure on the picture, assuming you know where the bombs are released from) and time since release. This equation should also involve the constant of acceleration due to gravity, but nothing else, and you can then solve the equation to give time as a function of distance.

Given the time since release of the first and last bomb found by this equation you can then proceed to figure out what the difference in time means for the trajectory of the two bombs as viewed from the ground where the plane speeds ahead and how this relates the their ground impact points.

Actually wait, the teacher added that we should assume that the distances between each bomb is 4m and therefore, I calculated that the bottom of the plane to the lowest bomb is a total distance of 44m if that matters. Is that relevant?

Actually wait, the teacher added that we should assume that the distances between each bomb is 4m and therefore, I calculated that the bottom of the plane to the lowest bomb is a total distance of 44m if that matters. Is that relevant?

Yes, that would be relevant, even though I must say the bombs in the picture clearly do not look like they are separated by the same distance.

Yes, that would be relevant, even though I must say the bombs in the picture clearly do not look like they are separated by the same distance.

This is true, though the question asks for the average distance between craters - presumably then, knowing the average distance between bombs will suffice.

To do that you could measure from the picture the distance between 12 bombs and divide by 12, since we know the length of the plane and the picture is pretty much an orthogonal side-view of the plane.

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This is true, though the question asks for the average distance between craters - presumably then, knowing the average distance between bombs will suffice.

To do that you could measure from the picture the distance between 12 bombs and divide by 12, since we know the length of the plane and the picture is pretty much an orthogonal side-view of the plane.

That would not be accurate. The division by 11 should be of either the ground distance between the impact point of the first and last bomb or of the release time difference between first and last bomb. If you divide the vertical distance between the first and last bomb on the picture by 11 and use that as the distance between two bombs you will get a number that is over 3 times bigger.

No, no, no, no, no. Stop. Stop everything. The OP, and ALL the responders, are completely on the wrong track. The teacher and tutor are idiots and should quit teaching.

Now that I've finished being melodramatic, take a closer look at the picture. Imagine you're on the plane, looking down at the bombs. What will be the separation between them when they hit the ground? What's their horizontal speed relative to the plane?

That would not be accurate. The division by 11 should be of either the ground distance between the impact point of the first and last bomb or of the release time difference between first and last bomb. If you divide the vertical distance between the first and last bomb on the picture by 11 and use that as the distance between two bombs you will get a number that is over 3 times bigger.

Whoops, of course. Sorry about that, it's nighttime here in the UK and for some reason I made the (obviously false) assumption that the distance of the bomb below the plane was proportional to how long ago it was released.

EDIT: ideasrule, the separation between bombs (horizonally, when they each hit the ground) will be the same as the distances the plane travels between dropping each bomb? I think filiplarsen had it correct, unless I've misunderstood the question?

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Actually wait, the teacher added that we should assume that the distances between each bomb is 4m and therefore, I calculated that the bottom of the plane to the lowest bomb is a total distance of 44m if that matters. Is that relevant?

Now I actually went and measured the distances on the picture. On the assumption that the vertical and horisontal scale of the picture are equal and the length of the plane is 60 m (even though wikipedia has the length of B52 at 48.5m), I get the lowest bomb to around 10 m below release point and the upper bomb around 2 m below. That is more than a factor 4 less than the 44m, so I would not recommend using 44m unless your teacher specifically has mentioned that you should use this exact number.

Now I actually went and measured the distances on the picture. On the assumption that the vertical and horisontal scale of the picture are equal and the length of the plane is 60 m (even though wikipedia has the length of B52 at 48.5m), I get the lowest bomb to around 10 m below release point and the upper bomb around 2 m below. That is more than a factor 4 less than the 44m, so I would not recommend using 44m unless your teacher specifically has mentioned that you should use this exact number.

Did you see my post? You DO NOT need to measure anything, assume anything, or calculate anything. This is a conceptual question with a simple answer.

Haha, wtf...

You have both teachers and tutors who cannot answer this problem?

I'd quit school, teach myself and stop paying money if the people who are teaching me this stuff couldn't answer...

First off; convert the plane's airspeed into metres per second.

Second; convert the plane's altitude (height) into metres.

Third, DON'T measure anything...

Fourth, stay away from wikipedia or government sites displaying B52 specs & measurements :p

(This reminds me of being 10 years old & measuring those geometric triangles in primary school geometry books trying to catch a teacher out :p).

Fifth, look at the picture. Think about it.

The bombs have all been dropped horizontally downwards in one go.
Also, they have been dropped from the middle of the plane.
You know the length of the plane.
You know they were dropped so gravity = 9.8m/s^2 is the acceleration.
(This will be consistent with all units if you followed my above advice).
What kind of equation would account for all of these factors?
The question asked for the average.
Why are people dividing by 11 when there are 12 bombs?

I'll admit that this is a good question - at first, but this is basic mechanics. The information about the length of the plane (and/or the fact that they are released in the middle of the plane) can be used to answer this in a variety of ways.

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Did you see my post? You DO NOT need to measure anything, assume anything, or calculate anything. This is a conceptual question with a simple answer.

I hope we do agree that the bombs in the picture do impact ground at different distances down range. If so, it should also be obvious that this distance must vary depending on the time between release, i.e. two bombs released one second apart will have a different impact distance compared to if they were released 1 minute apart. So if you want to give a numerical answer based on the observed facts, you must indeed understand the physics, make your assumptions, measure something and then calculate. Sure, if you want to give an algebraic answer (average distance between bombs expressed in terms of quantities whose value can be either measured, estimated or assumed) you can indeed skip the actual measuring and the actual numerical calculation.

If you think otherwise then feel free to present your argument. Or we should perhaps wait a bit in order not to spoil the "fun" for OP and then each present our solution?

Why are people dividing by 11 when there are 12 bombs?

If there were 2 bombs would you then be dividing by 2? Or think about it the other way around. If you have n points distributed with equal distance d between each other along a line, what is then the total distance between the first and the last item expressed in terms of d and n?

Two bombs released one second apart will fall at a fixed distance from one another.

The third bomb will fall the same distance away from the thirs as the second did from the first (assuming they are all dropped in equal intervals of time - as the question explicitly asks you to assume due to the word "average").

You can get very specific by using the facts I listed in my first post (plane size, bombs being dropped from the center of the plane) & these pieces of information weren't included in the question just randomly...

Yeah I love the idea of presenting an answer once the OP gets his version lol

If there were 2 bombs would you then be dividing by 2? Or think about it the other way around. If you have n points distributed with equal distance d between each other along a line, what is then the total distance between the first and the last item expressed in terms of d and n?

Yeah that makes sense, I was thinking of it starting from the origin for some reason.

I have a nagging suspicion that I can make it work by dividing by 12 still lol, I'll mull it over & see if it can be incorporated :tongue:

This problem can be easily solved if one is a wee bit organized in one's thoughts. Clearly, the distance between craters is the distance traveled by the plane between successive releases. Let

t0 = the time between successive releases, assumed the same for all bombs.

Then the vertical positions of the bombs will be

y1 = (1/2)g(1*t0)2
y2 = (1/2)g(2*t0)2
...
y12 = (1/2)g(12*t0)2

Using the body of the plane as your scale, find distance y12 from the photograph and from this find t0. Once you know t0, you can easily find the distance between craters.

BTW, if the bombs hit at different horizontal distances, how come they are directly below each other in the photograph? That's a conceptual question to ponder.

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BTW, if the bombs hit at different horizontal distances, how come they are directly below each other in the photograph? That's a conceptual question to ponder.

It may be the total lack of sleep but are you asking or trying to encourage the OP lol, there's an answer even though the picture looks a bit odd.

It may be the total lack of sleep but are you asking or trying to encourage the OP lol, there's an answer even though the picture looks a bit odd.
I am not asking, just pointing out what would have been obvious to Galileo.