# Homework Help: Firecracker explodes

1. Dec 6, 2013

### Panphobia

1. The problem statement, all variables and given/known data
A firecracker explodes at a height of 100 m above level ground. It breaks into a large number of very small fragments, which move in every direction; all of them have the same initial velocity of 20 m/s.

We assume that the fragments move without air resistance, in a uniform gravitational field with g=9.81 m/s^2.

Find the volume (in m^3) of the region through which the fragments move before reaching the ground. Give your answer rounded to four decimal places.

3. The attempt at a solution
So this is a projecteuler question I thought I could with pen and paper but it seems much more complicated than I thought. The initial shape of the explosion is a sphere but since there is gravity, the shape kind of distorts. What kind of math would I need to know to solve this question?

2. Dec 6, 2013

### phinds

One thing you might think about is that each piece will follow a ballistic trajectory. What kind of curve is that?

3. Dec 6, 2013

Parabolic?

4. Dec 6, 2013

### phinds

Yes, so what does that tell you about the envelope of the enclosed volume?

5. Dec 6, 2013

### collinsmark

That's right.

Let's think about that. Are you sure?

Let's investigate that idea for a moment.

Consider two objects on a frictionless plane. Suppose they shoot away from each other at a constant velocity, each with a constant velocity v0, in opposite directions (If it helps, one has a velocity of +v0 and the other of -v0). What is their distance from each other after time t?

Now repeat but with the objects in the sky, undergoing acceleration which is perpendicular to their initial velocity. I think it's easy to see that their relative distance is the same after time t as it was before (not the distance to the ground, but the distance to each other). The acceleration is perpendicular to their initial velocities so it doesn't change their relative separation as a function of time, t.

Now calculate their relative distance from each other (not the ground) after time t, if one shoots straight up with initial velocity v0, and the other straight down with initial velocity -v0. Is their distance from each other*, after time t, any different than it was before?

*(Not the distance to the ground, but the distance from each other. That's an important distinction when determining the shape)

Last edited: Dec 6, 2013
6. Dec 7, 2013

### phinds

collinsmark, correct me if I'm wrong, but you seem to be arguing that the envelope would be spherical, which it most certainly would not. A cross section of the envelope, parallel to the ground, will at any given point be a circle, but that in no way implies that the envelope will be a spherical since if it were that would imply that a VERTICAL cross section would be a spherical but it won't.

7. Dec 7, 2013

### Panphobia

See thats what I thought ^ also is this volume the instant that the first fragment touches the ground?

8. Dec 7, 2013

### collinsmark

I was hoping that Panphobia would figure this out, but I'll have to interject.

The envelope will be spherical.

The initial conditions are that all particles start with a spherical distribution of velocities (constant magnitude, but spherically symmetrical directions). And air resistance and any other form of friction can be ignored.

The important point is that all particles accelerate with the same magnitude, same direction. All particles have zero relative acceleration with respect to each other.

First let's take the horizontal cross section for reference. The variable d is the distance between two particles moving in opposite directions. Time t is such that t = 0 at the moment of the explosion.
$$d = (v_0 t) - (-v_0t) = 2v_0 t$$
Now look at the vertical distance between two particles (one going straight up, the other straight down).
$$d = \left( v_0 t - \frac{1}{2}gt^2 \right) - \left(- v_0 t - \frac{1}{2}gt^2 \right)$$
$$= v_0 t - \frac{1}{2}gt^2 + v_0 t +\frac{1}{2}gt^2$$
$$= v_0 t + v_0 t - \frac{1}{2}gt^2 + \frac{1}{2}gt^2$$
$$= 2 v_0 t$$
'Same diameter in every direction. Do you see how that makes sense?

9. Dec 7, 2013

### collinsmark

That's the way I interpret it, yes. Find the volume at the moment that the first fragment touches the ground.

10. Dec 7, 2013

### TSny

I think they are asking for the volume of the largest region of space such that each point in the region has at least one fragment pass through the point (by the time all of the fragments have reached the ground). But I could be wrong.

11. Dec 7, 2013

### Panphobia

I don't know exactly what you guys mean by envelope, but if it is a sphere throughout the entire process like you said collin, does that mean I can take the distance between the bottommost particle and topmost particle and call that the diameter, then get the volume from there?

12. Dec 7, 2013

### collinsmark

Ah, that would make a difference. I guess it depends on how one is to interpret, "Find the volume (in m^3) of the region through which the fragments move before reaching the ground."

Does that really mean "just before any fragment reaches the ground," or does it mean "up to the point in time that all particles reach the ground"?

[Previously I interpreted the problem as meaning the volume at the moment that the first particle reaches the ground. But now I'm not so sure. :uhh:]

13. Dec 7, 2013

### TSny

It might help to consider the following question. Let the y-axis run perpendicular to the ground and pass through the point of explosion. Imagine a very tall pole erected perpendicular to the ground at a horizontal distance x from the y-axis. What is the maximum height on this pole that gets hit by a fragment of the explosion?

14. Dec 7, 2013

### TSny

Yes, it's a matter of interpretation. I lean toward the "all particles reach the ground" interpretation.

15. Dec 7, 2013

### collinsmark

Panphobia,

Is this problem for a calculus based physics class? (or is calculus not a requirement?)

If it's not a calculus based class, have you been given an "optimal range equation" in your coursework that you can plug in difference in height and initial velocity and you get the "maximum" horizontal distance of the projectile?
(Not a simple range equation that gives you the x-distance traveled for a given angle θ and v0, but one that gives you either the maximum distance or the optimal angle based on height Δh, and initial velocity v0?)

16. Dec 7, 2013

### Panphobia

No this is a question for fun that is turning out to be harder than I thought

17. Dec 7, 2013

### collinsmark

If we assume the interpretation of "the total volume that any fragment ever passes through from the moment of the explosion until all particles reach the ground," interpretation, you could solve this problem with calculus.

You could derive the range equation that gives horizontal x-distance as a function of height difference Δh, angle θ, gravitational acceleration g and initial velocity v0. Alternately, you could just look it up. Make sure to use the version that is dependent on height.

Take the derivative, with respect θ, of that range equation and set the result equal to zero. Solve for θ, plug that θ back into your original range equation and solve for the horizontal distance x, and now you have an equation that gives you the maximum horizontal distance from the "center pole" for any given height.

Turn that into a volume using a technique called "volume by revolution." That technique involves some calculus. You divide the shape into many thin, horizontal disks. Then integrate the disks from the ground to the maximum height.

[Edit: I've looked at some of the math and it doesn't look very friendly (though nothing impossible). It might help if you have Mathematica or maybe you can use WolframAlpha.]

Last edited: Dec 7, 2013
18. Dec 7, 2013

### Panphobia

This is what I was thinking, get the maximum of this shape, and then get the x intercept of one half of a slice, and kind of rotate it around the z axis to get the volume, do you think that could work? This might be the exact same thing you said, but I didn't understand haha. I got
X = 20*cosθ*t
0 = 100 + 20*sinθ*t - 4.905*t^2

After that I really don't know what to do because 3 unknowns and only 2 equations.

Last edited: Dec 7, 2013
19. Dec 7, 2013

### phinds

My statement about it not being spherical was, of course, based on the interpretation of all particles hitting the ground.

If you take the limiting case of the height of the explosion approaching infinity above an infinite plane that has uniform downward gravity, then the envelope clearly approaches being 1/2 of a paraboloid. What I cannot figure out is whether or not it is a cut-off paraboloid in general. Clearly it is close, but "close" will not get the right answer if it's "close but no cigar".

EDIT: oops ... clearly "1/2 of a paraboloid" is not a meaningful statement. I should have just said cut-off paraboloid

20. Dec 7, 2013

### TSny

From

x = vo(cosθ)t
y = yo + vo(sinθ)t - (1/2)gt2

you can eliminate t and get the trajectory y as a function of x for a fragment thrown out at angle θ.

As mentioned in post #13, you can now fix x and vary θ to maximize y at a fixed distance x. I believe this should give you ymax(x) which will be your envelope curve that you can then rotate around the y axis to generate the surface that encloses the volume above the ground.

Carrying this through and finding the volume did not seem too difficult, unless I overlooked something.

21. Dec 7, 2013

### phinds

Possibly I am missing something but I don't see how could possibly work. It clearly would work if none of the pieces ever went above the point of the explosion but that's not the case.

22. Dec 7, 2013

### TSny

I don't follow. After solving the equations for the envelope curve, I made the attached animation which shows that it does appear to envelop the trajectories.

#### Attached Files:

• ###### trajectories.gif
File size:
70.3 KB
Views:
149
23. Dec 7, 2013

### collinsmark

Brilliant!

And here I was trying to maximize x over θ, which gives you the same thing but involves a lot more algebra.

Maximizing y over θ is easier since the equations are initially set up for that.

By the way, for what it's worth, when doing the calculations I found a trig identity very useful (which can be derived from 1 = sin2θ + cos2θ):

$$\frac{1}{\cos^2 \theta} = 1 + \tan^2 \theta$$
Identities like that allow you to avoid having trig functions of other inverse trig functions.

Last edited: Dec 7, 2013
24. Dec 7, 2013

### phinds

Very cool. I clearly mis-thought what happened to the envelope as the pieces went above the point of the explosion.

What tool did you use to make that animation?

25. Dec 7, 2013

### TSny

I used the "Manipulate" command in Mathematica to create the animation. I exported the animation as an avi file and then used another program to convert the file to an animated gif.