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Name: David
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In the real world, (not the physics world of no air resistance), how do I calculate the speed of a bullet that was shot straight into the air when it returns to Earth. For example, say a rifle had a muzzle velocity of an M-16, what would the bullet's speed when it comes straight back down. On the news one sees celebrations in the Middle East of shooting into the air. Just how dangerous is that?

The upper limit is of course as you point out (no air resistance). Including air resistance is a much more complex calculation because it depends upon many other factors -- air density as a function of height, the shape of the bullet, the rotational speed of the bullet, whether the bullet is wobbling or tumbling in the way down Really a complicated mess to calculate, but for the moment DO ignore air resistance. A rule of thumb in ALL physics/chemistry/engineering is do an order of magnitude calculation to see what, if any, more complicated calculation may be necessary -- but always carry out "reality checks" to make sure you are on track. Let us consider a rifle bullet vs. a hailstone (spherical).

RIFLE BULLET: A typical muzzle velocity of a rifle (Google search) is 3000 ft/sec = 1000 m/sec (notice I am rounding here because we are just looking for where the decimal falls. A typical bullet mass is 120 grains [weird units, but 1 grain = 0.065 gm] = 7.8 gm = 10 gm (close enough). Now a "reality check". The density of lead is 11.4 gm/cm^3 = 10 gm/cm^3 (close enough). So the volume of the bullet is: volume = mass / density = 10 gm / 10 (gm/cm^3) = 1 cm^3. That is probably pretty conservative, but OK for an order of magnitude. Remember the shell casing does not count -- only the projectile. In the absence of air and a perfect world, kinetic energy is conserved, so the bullet weighing 10 gm will hit the ground after a vertical trajectory at a speed of 1000 m/sec. OUCH!!! Let us calculate the energy. From the muzzle velocity (1000 m/sec) and the bullet mass (10 gm = 10^-2 kg) and K.E. = 1/2 m(v)^2 we get 1/2 (10^-2)*(1000)^2 = 0.5*10^-2+6 = 5x10^3 = 5000 Joules. HAILSTONE: A spherical hailstone weighing 10 gm has a volume of 10 cm^3 since the density is 1 gm/cm^3. REALITY CHECK: The volume = 10 cm^3 = 4/3 *pi* r^3. So r^3 = 2.4 cm^3 or a radius of r = 1.3 cm or a diameter of 2.6 cm (a fairly nominal hailstone -- about an inch in diameter).

The potential energy of a hailstone weighing 10 gm = 10^-2 kg

falling from 10 km is: P.E. = m*g*h =

10^-2 * 9.8 * 10,000 (about 10^-2+1+4 = 10^3 = 1000 Joules). Its velocity

assuming complete conversion of the P.E. to K.E. = 1/2*m*(v)^2

gives: 1000 = 1/2x10^-2*(v)^2 or v^2 = 200,000 m^2/s^2 or about 450 m / sec

Compared to 1000 m/sec for the bullet.

Now if you want to refine the estimate further a fair assumption would be to assume that air resistance would be proportional to the cross sectional area of the object and the time of flight. That is, the longer the object is in the air the greater will be the drag from the atmosphere until the projectile reaches its maximum terminal velocity. Even without doing the calculation, assuming the bullet is a cylinder (you can vary the length / diameter ratio) and the hailstone is spherical, the air resistance will be much less for the bullet than for the hailstone for two reasons -- cross sectional area, and time of flight. This also gets a bit messy because the bullet experiences drag both going up and down, but the hailstone only experiences drag on the way down. Of course, there are other complicating factors that have been ignored. A big one is that the speed of the hailstone will depend upon whether it is falling in an up-draft, or is being accelerated by being in a down-draft.

As to danger there is no question that getting hit by such a spent projectile could be lethal. Actually getting hit by a 1 inch hailstone would be very unpleasant and possibly also fatal.

I found both a better explanation and a plug-in algorithm, at least for spherical objects launched vertically, on the great website.

The drag force depends upon the square of the velocity, a shape dependent drag factor "C", the cross sectional area, and the density of air. The key point is that it is NOT negligible. Depending upon the values of the inputs the projectile may / may not attain a constant terminal velocity.

The inputs assumptions are: a spherical object, standard gravity constant for Earth, constant atmospheric density, laminar air flow, no spinning or wobbling (not an issue for a sphere), constant temperature, and of course vertical trajectory. The algorithm then cues the user for the density of air, the shape-dependent drag coefficient, C, the radius of the sphere, the density (or mass) of the spherical projectile, the initial velocity. The default values: are 1.29 kg/m^3 for the density of air, C = 0.5 (C can vary from about 0.1 to 2.0), but these can be changed. The algorithm outputs are: terminal velocity, the velocity at any intermediate height on the way up, the peak height, the time to achieve peak height, the velocity at any intermediate height on the way down. This allows you to find the velocity upon impact.

For comparison, I chose lead (density = 11.3 gm/cm^2 (47.3 gm)) and ice (density = 0.9 gm/cm^2 (3.8 gm)), an initial velocity in both cases of 3000 m/sec (of course, this could be changed). I left the other parameters at their default values. The outputs were:

For lead: peak height = 1771.5 m, peak time = 10.7 sec, terminal velocity = 67.7 m/ sec, velocity upon impact = 67.6 m/sec, time to impact = 31 sec. Note: the lead projectile reached its terminal velocity before impact.

For ice (hail): peak height = 188.2 m, peak time = 3.0 sec, terminal velocity = 19.1 m/sec, velocity upon impact = 19.1 m/sec, time to impact = 11.2 sec. Note: the hail stone also reached its terminal velocity prior to impact. The momentum upon impact for lead and ice would be: 3.2 kg*m and 0.07 kg*m, respectively. In either case you would take a pretty nasty hit with either projectile.

The algorithm is a good teaching tool because it takes the pain out of the number crunching. This allows the physics student to carry out all manner of thought experiments without getting brain dead from the manipulations. The bottom line is initial velocity and air friction are very important factors. I think that the non-vertical trajectory must be solved numerically, because the gravitational force and air drag force are not co-linear, but I am not sure of that.

Vince Calder

Setting the force of air resistance (F=rCAv^2/2) equal to the weight of a piece of hail, I get a terminal velocity of 57 m/s. Note that this is an engineering estimate not rising to the level of Newton's Laws. There is a caution that for more than 500 meters, this relationship may not hold true. But assuming it does hold true, we continue. For example for the effective area, I take the cross sectional area of a 1/4" diameter sphere and use a C factor suitable for spheres, 1.2.

In the equation, r is the density of air (1.3 kg/m^3), is a fudge factor (I took 1/3), A is the cross sectional area (3.9E-5 m^2) and v is the speed in m/s. For the hail I got a mass of 2.2E-4 kg (around 0.008 oz). Equating the air resistance to the weight of the hail gives 57 m/s for the terminal velocity.

If one stops the hail with constant acceleration in 1 cm, a force of 35.7 N (of 8.7 lb) is necessary; to stop in a millimeter, 87 lb. , These do not seem to be fatal forces. Incidentally, I find it very difficult to make these calculations without error! I always try to do them in at least two different ways as a check, but...

As for the rifle recoil: If you fire a 30 gm bullet at 2800 m/s (numbers taken off a web site), a 10 lb (4.5 kg) rifle recoils at 3.8 m/s. To stop the rifle in 2 in, a force of 146 lb is required. Wow! Barely conceivable.

At another site ( I found a 0.458 Win. Mag. which fires a 500 gr (thats grains) bullets at 2100 ft/s. This is a 9 lb rifle and the recoil is quoted as 21.1 ft/s. Matching the momentum of the rifle to the momentum of the bullet, I get a recoil speed of 16.6 ft/s. I have no understanding of why conservation of momentum is not the same for me as for them. Especially as their amount of non-conservation varies among the rifles. Anyway, to stop the recoil of this 4.1 kg rifle in 2 in (0.05 m), the acceleration must be 260 m/s^2 and the force is 1066 N or 240 lb. I guess I just do not understand rifles; it seems to me that force could knock someone down. Maybe it does??? Of course if the marksman stops the gun in 8 in, the required force is only 60 lb...

Best, Dick Plano, Professor of Physics emeritus, Rutgers University

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