Physic: bomb trajectory calculation

Hi

I tried to solve this today and well I have not done any math or physics since
over 15 years and to my frustration noticed that I lost it all.

Can you help me to solve this?

Question:

Can the object hit a target which is 39 km (39 000 m) away?

Object information:

weight 450 kg
height 15 km (15 000 m)
speed 1593 km/h (Mach 1.5)

Yes, we can help, but we need a bit more information.

I guess “the object” is some kind of projectile. Wich is thrown or left to fall freely from 15.000 m above the target level.

I guess also that the speed is the speed of the projectile in the moment it is thrown or left to fall. But the difference is important. HOW is the projectile thrown? Is it thrown from a weapon that can shoot in any angle? Is it left to fall from a plane flying paralell to the ground? Is it shooted with a constant angle? (angle relative to the horizontal ground).

A painting of the problem will be very helpful :slight_smile:

ok sorry

horizontal bomb drop

plane speed and thus bomb forward speed is 1.5 mach

dropped from a height of 15 km.

All I need to know is the distance it can fly.

To my surprise I did not find any calculator online since it seem
that this is such a common physics equation.

thus the bomb trajectory should look like this:
http://www.math.uiowa.edu/~stroyan/CTLC3rdEd/ProjectsOldCD/estroyan/epsgif/323.gif

cool drawing, eh? :slight_smile:

Great drawing ;-).

Give me some minutes to calculate. Coming back in a moment…

Calculated.

Result: No, it can’t reach such distance. Maximun distance is 24,482.79 meters (about 24.5 km).

Further explanation coming in some minutres (just in case anyone else want to validate the process :))

Alright. I will try to make this as simple as I can. Do you realize that we are on planet that is revolving at somewhere around (I think) 1800 miles per hour? So what does this mean? Okay quite simply, if you should shoot a gun straight up in the air, I will assure you the the bullet will not even land close to you, not even close to the poles. This was one reason for inventing computers. Then they found blender! It is more fun and a lot less hazardous to anyone’s health!

ah six grade math.

the calculation is easy:

First we need to calculate the time it takes for the object to hit the ground. This without taking the curvature of the earth into the calculation.

basic formula is x(distance) = 1/2 * g * t^2

g is 9.81 average iirc so the result should be roughly 39.10 sec.’

Then if we calculate the horizontal distance with the formula x = v*t then we become something of around 62 290 m

Of course this does not take friction into account or the curvature of the earth or the fact that there is lesser ‘gravity’ 15 km from the surface.

But bombs are pretty aerodynamic so my guess is that it should be able to travel 39 000 m.

The Paperclip.

Ok, lets separate the vertical and horizontal movements of the bomb.

In the vertical direction, the bomb is left to fall freely from 15 km, with Earth’s gravity accelerating it towards the ground. We can therefore calculate how long will it take to reach the ground using the formula:

s = s0 + v0·t + 1/2 · a · (t*t)

Where s is the position of the bomb, s0 is the initial position, v0 is the initial speed (in the vertical direction, that is, zero), a is the acceleration (this is g = 9.8 m/s*s) and t is time. So, in our case:

0 = 15000 + 0·t + 1/2 · (-9.8) · (t*t)

Solving the second degree equation you get t = 55.328 s, wich is the time elapsed from the moment the bomb is dropped to the moment it reaches the ground.

Now, after that time, and considering that the bomb is flying in the horizontal direction at a constant speed of 1593 km/h (ie 442,5 m/s), the bomb have traveled

s = s0 + v0 · t

Wich, in our case is

s = 0 + 442.5 · 55.328 = 24482,7876 m.

As you can see, the weight of the bomb is not important at all.

NOTE: We have ignored other forces that may be actuating on the bomb, specially air friction. In any case, to take air friction into acount will make the bomb travel even LESS than 24.5 km so, by any means can it reach 39 km.

You will need to:

  • Fly faster, or
  • Release the bomb closer to the target, or
  • Go bombing to another planet, where gravity is less intense.

Good luck!

lol I found the answer in a document which analyses the bombing in the movie pearl harbor.

But no help at “how stuff works” …

Sw= root of ((height*2)/gravity) * Velocity

1593 KM/h = 442.5 m/s

root of (30000/9.81) * 442.5 = 24470.3059

Impressive how far a thrown object can fly.

I am curious about the gravity.
Well any objects accelerates until it reaches its terminal velocity.
Also wind drag should slow down the object a little, when not much as I assume a bomb might be
shaped aerodynamically to minimize that.

Paperclip: Check your results. I guess you mixed m/s with km/h, so you got funny results…

pixelmass: Are you sure? I can’t see the Earth moving at all…

What I mean is: Yes, the Earth may be moving, but this is not important. I am moving along with it. And the projectiles I throw are moving along with it as well so, for us (me, you and the projectiles) there is no movement at all.

Is the same thing that happens if you let an apple drop while flying in a plane. Regardless of the plane speed, the apple will fall right on your feet, and won’t travel to the plane’s tail!

@NoeOM: I forgot to divide by 2. I forgot it out of ‘quickness’. Immediatly spotted it when I saw your explanation. And I forgot to turn km/h into m/s. So far for trying to be quick :o

Paperclip: Yep, and I realize your mistake because I had made exactly the same few minutes ago hehe.

The only difference is that I double checked the solution :wink:

Thinking about it, it can reach longer.

If the bomb has WINGS or any other kind of sustentation, it can reach longer than it will if freely falling.

Paperclip the 6th grade math is wrong :wink:
But I made the same mistake as well.
You need to divide km/h through 2.6 to get m/s.

In free fall the object also has a forward speed(earth rotation speed)
in case it would be attached to a structure resting on the ground.

I assume it is more than minimal but on the other hand rockets
are launched from the equator because of the higher speed there
and less fuel needed. Earth rotation is 1,674.4 km/h.

lol

that was an epic km/h to m/s conversion group fail

Sheesh, I dunno, I only studied this for several years. Perhaps I should go back to school? After all, what can be more important than insuring that you hit your target? Well, that is if you have one.
:RocknRoll:

Hmm…

The G-20 summit site is approximately 39km from the US border if you account for the distance it would take a plane traveling at mach 1.5 to change direction and not ‘invade’ Canada.

Coincidence, perchance?

Whoops, I forgot some links for you…
http://www.google.com/search?q=ballisic+trajectories+math&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

That’s not true, the earth is rotating, and the further away from the earth you are, the higher the velocity due to the rotation. So If you fire a bullet straight up in the air it will travel west (not east as you might think). I have the math for this written down, but I won’t be home until Sunday evening, I will post it then unless someone beats me to it.