# Deceleration of a bullet [closed]

Taking into account gravity, air resistance, and wind velocity and direction. (And also temperature if it's actually relevant.) I know the muzzle velocity, the total distance the bullet traveled, its diameter, mass, length, and the angle from which it was fired at. Now, I do not know its drag coefficient, so how do I calculate it? (The muzzle velocity is supersonic.)

The question is, what is the velocity of the bullet by the time it traveled the total distance? What are the formulae that I would need to use to calculate that?

Much obliged.

## closed as off-topic by ACuriousMind♦, Kyle Kanos, Bill N, Brian Moths, user36790 Oct 3 '15 at 2:43

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• "what is the velocity of the bullet by the time it traveled the distance?"-Which distance? Total distance or something else? – SchrodingersCat Oct 2 '15 at 14:13
• Correct. I apologize, I'll edit that in. – 1234567 Oct 2 '15 at 14:17
• I'm voting to close this question as off-topic because the OP has not shown any effort to research projectiles subject to air resistance. Several intermediate physics and engineering textbooks give examples of how to handle this problem. – Bill N Oct 2 '15 at 15:39

Let's start off by writing down all the forces the bullet experiences from the moment it leaves the muzzle all the way to the moment it lands on the ground:

1. Gravity:

$\vec{F}_g=- m g \hat{y}$

where $\hat{y}$ is a unit vector pointing upwards, and that's why there's a minus sign there, to make it point downwards.

1. Drag force:

$\vec{F}_d =- \frac{1}{2} \rho v^2 C_D A \hat{v}$

which is in the opposite direction of the bullet velocity. $\rho$ is the density of air, $v$ is the speed of the bullet, $A$ is the cross section of the bullet, and $C_D$ is the drag coefficient. Similarly, $\hat{v}$ is a unit vector pointing in the direction of the velocity of the bullet.

Now that you have all the forces, you just need to plug them in Newton's equation of motion to calculate the trajectory of the bullet:

$\vec{F}_{tot} = m \vec{a}$

where $\vec{F}_{tot} = \vec{F}_g + \vec{F}_d$, and $\vec{a}$ is the acceleration vector. $m$ is the mass of the bullet.

Now all you need to do is solve this differential equation. As far as I know this differential equation cannot be analytically solved (edit: well, as @Gert has mentioned in a comment below, this is not true. Reference. Nonetheless, the rest of this answer is still accurate as a numerical approach), therefore, I would use the shooting method as the simplest approach. I would start with a drag coefficient close to that of the air at let's say 25 degrees Celsius, and calculate where the bullet lands. If it was further than the given distance, I would increase the drag coefficient and if it was shorter, I would decrease the drag coefficient. You can keep doing this until you get the desired accuracy.

• Simply marvelous, sir. Excellent. But one last question if it's not too much trouble, how do I find the speed of the bullet if I don't have a time-frame? – 1234567 Oct 2 '15 at 15:07
• That's a good start, but incomplete. For pretty much anything more powerful than a .22 short, the bullet will exit the muzzle supersonic, with a different drag equation, or at least a different drag coefficient. Note that in the supersonic regime Cd varies wildly with Mach number. – WhatRoughBeast Oct 2 '15 at 15:17
• In order to numerically solve this differential equation for a given drag coefficient, you almost always have to calculate the velocity at each time step in addition to the new position vector (there are many methods available on the internet, e.g. velocity verlet). Thus, you only have to know when it lands. At the moment of landing, the $y$ component of the position vector goes negative, and this is actually your criterion for stopping the simulation. The velocity at that point is the velocity you are looking for. – Mehrdad Oct 2 '15 at 15:19
• @Mehrdad: "As far as I know this differential equation cannot be analytically solved". That is incorrect. You can find the analytical solution here: en.wikipedia.org/wiki/…. Scroll down to the case of $F_{air}=-kv^2$. – Gert Oct 2 '15 at 15:36
• @Gert: Sweet, I didn't know that. TBH, I was told so back in university and was too lazy to check it out myself. Thanks for that. – Mehrdad Oct 2 '15 at 15:47

In cases with a reasonably high launch angle, drag will dominate, so the bullet will land with a terminal velocity defined by its shape alone, because the bullet will have lost nearly all its horizontal speed due to drag. It then falls under the force of gravity and won't exceed terminal velocity.

Obviously if your angle is, say $-90^{\circ}$ , The final velocity is just initial velocity minus a tiny drag loss. As the angle increases, there's a longer and longer drag time, up to the minimum angle at which my first paragraph applies. However, if you don't know the drag coefficient you can't calculate that angle.

(I guess this should be a comment, but it got kind of long)

• Hmmm...I like where this is going. The angle is −45∘. What do you say, good sir? – 1234567 Oct 2 '15 at 14:38

The question is, what is the velocity of the bullet by the time it traveled the total distance? answer Zero m/sec. It came to a stop.

• Jokes belong in comments, not answers – Carl Witthoft Oct 2 '15 at 17:20
• yeah better delete this joke answer before it gets downvoted and you loose reputation – Santropedro Apr 10 '18 at 18:28