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I read recently that if you hold a bullet in one hand and a pistol in the other, both hands at the same height, and subsequently fired the pistol at the same time as dropping the bullet, both bullets would hit the ground at the same time (assuming the fired bullet encounters no obstacles).

Is this true, and if so can you explain the reasoning behind it with as few equations as possible? I assume it is essentially that gravity is unaffected by momentum, but it just sounds counter-intuitive.

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    $\begingroup$ See also this MythBusters episode youtube.com/watch?v=D9wQVIEdKh8 $\endgroup$ – Qmechanic Jun 4 '11 at 20:43
  • $\begingroup$ In vacuum, yes. Otherwise it probably takes a little longer to reach the ground (think in terms of aerodynamics) $\endgroup$ – Diego Jun 4 '11 at 21:13
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    $\begingroup$ Actually, assuming ideal conditions, and no aerodynamic effects, the dropped bullet should hit the ground first because of the curvature of the Earth. The ground curves away from the bullet that is fired from the gun. $\endgroup$ – Solomon Slow Jul 7 '17 at 0:41
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    $\begingroup$ The question should specify that the pistol is fired horizontally. $\endgroup$ – Pere Nov 13 '18 at 15:20
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This is indeed true. The easiest way to think about it is in planes. You have horizontal and vertical planes for velocity and acceleration. For both bullets, there is initially no vertical velocity and the only net force acting upon them is gravity. You would then expect them to act similarly in the vertical plane, which is why they both hit the ground at the same time. There is nothing in the forward motion of the shot bullet that counteracts the pull downward from gravity (such as lift). I hope that explanation isn't too vague and helps you understand better what is going on.

EDIT: I should specify that this only holds true for ideal situations (being in a vacuum, no air resistance, etc.)

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    $\begingroup$ I would say "directions" rather than "planes". $\endgroup$ – David Z Jun 4 '11 at 22:48
  • $\begingroup$ I had originally put vectors, but wanted to keep it as basic as possible just in case that was unfamiliar territory $\endgroup$ – slev Jun 4 '11 at 23:09
  • $\begingroup$ That makes sense. I assume this also assumes no recoil and no arc imparted by rifling as in the comments to Mark Eichenlaub's answer? In other words assuming a perfectly flat trajectory for the fired bullet. $\endgroup$ – ElendilTheTall Jun 5 '11 at 16:57
  • $\begingroup$ Yes, it assumes no recoil. Mark's answer is much more complete in terms of what would happen in the world we live in. I interpreted your question to be asking in terms of Newtonian physics, in which case my answer holds true. But in the actual world, those rules only hold true as approximations since Coriolis forces and air resistance will impact the complexity of the equations governing the bullets' actions $\endgroup$ – slev Jun 5 '11 at 17:32
  • $\begingroup$ Accepted this as it is the simplest explanation that the man on the street (read 'me') would understand. I'd accept Mark's as well if possible, but equations bring me out in hives. $\endgroup$ – ElendilTheTall Jun 6 '11 at 8:30
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The bullets don't hit the ground at the same time exactly because it is very difficult to fire horizontally, there is air resistance to account for, the ground may be sloped, there are Coriolis forces, etc.

Usually, when people refer to this phenomenon, they're referencing the principle of Galilean relativity.

You can read the famous excerpt from Galileo's book here. He considered sailing on a smoothly-sailing ship, and noted that if you're in the hull, you don't notice the ship's motion in any physics experiment.

Specifically, things seem to fall completely normally when you're in a moving ship. Someone standing on the shore watching would think that objects in the ship are falling in parabolic trajectories with constant forward speed equal to the ship's speed, but someone in the ship thinks thinks they fall straight down.

Now imagine two bullets. The first is dropped by someone on the ship. The second is fired from a gun on the shore, but fired at exactly the same speed as the ship's speed (it's either a slow bullet or a fast ship, or both). The bullets start out with exactly the same trajectory. They have no way of knowing whether they're supposed to keep up with the ship or the shore or what - they only know their location and velocity. So the two bullets must fall in the same way. By Galilean relativity, the bullet on the ship falls in the same amount of time as a bullet that's simply dropped, so the bullet fired from the gun falls in the same amount of time as a bullet that's simply dropped, too.

One way to view the math is as a consequence of Newton's laws of motion, which state

$$F = m \frac{\textrm{d}v}{\textrm{d}t}$$

If we transfer reference frames from the shore (frame S) to the ship (frame S'), we're transforming the velocities according to

$$v' = v + v_s$$

with $v_s$ the constant speed of the ship.

Then

$$F' = m \frac{\textrm{d}v'}{\textrm{d}t} = m \frac{\textrm{d}v}{\textrm{d}t} = F$$

so physics appears to work the same way in both frames (this assumes that mass is invariant).

However, the relativity principle is deeper than Newton's laws, and holds in theories of physics that are more fundamental than Newtonian mechanics, so it also makes sense to take relativity as more of a starting point than a consequence.

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  • $\begingroup$ I am confused by the statement: "it is very difficult to fire horizontally". What does that mean? $\endgroup$ – Bill Slugg Jun 5 '11 at 1:44
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    $\begingroup$ @Bill Slugg For most guns there is going to be kick back and human error in steadily holding it level. Since bullets travel very fast compared to an object falling from those heights even a slight tilt imparts a vertical component to bullet's velocity. A handgun for example would have significant error. $\endgroup$ – David Jun 5 '11 at 5:46
  • $\begingroup$ There is a much stronger influence on real bullets: rotation due to rifling ( en.wikipedia.org/wiki/Rifling ) aerodynamics of this makes some uplift for the bullet. To avoid it, one had to use some unrifled gun or a perfectly spherical bullet. $\endgroup$ – Georg Jun 5 '11 at 12:49
  • $\begingroup$ I understand now. I was assuming an idealized case. $\endgroup$ – Bill Slugg Jun 5 '11 at 13:49
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Mark came the closest in his answer, but I want to address some of the deviations from idealism in greater detail. People have mentioned the angle the gun is at and air resistance, so I'll try to touch on both.

You have a height $h$, say the velocity of the bullet is $\vec{V_b} = < V_{bx}, V_{by}>$ (add $t$ dependence when necessary). The time it takes to fall will be $\sqrt{2h/g}$ in a vacuum based on ideal Newtonian kinematics. The idea in the frictionless world is that the horizontal velocity of the bullet remains $V_p$ and the resultant velocity over time will be a combination of that and the vertical velocity caused exclusively by gravity, $g t$. I will use $<x, y>$ vector notation, and I'll just use a big vector arrow to indicate what is shorthand for a vector. For the other ball that just falls, I'll denote it $\vec{V_f} = <V_{fx},V_{fy}>$.

$$\vec{V_b}(t) = < V_p , -g t>$$ $$\vec{V_f}(t) = < 0, -g t>$$

Next, air resistance (drag) is usually taken to be some proportional relationship to velocity, in the opposite direction of the velocity vector. I'll take that relationship to be the common form of a constant times the magnitude of velocity squared (linearly $a_b = -C V^2$). Then I'll write the full differential equations for the problem... in some fairly extreme vector shorthand notation. Take $r$ to be the position.

$$\vec{a_b}(t) = -\frac{\vec{V_b}(t)}{|\vec{V_b}(t)|} C |\vec{V_b}(t)|^n = -\vec{V_b}(t) C |\vec{V_b}(t)|^{2} $$ $$ \vec{r_b}' = \vec{V_b}$$ $$ \vec{V_b}' = \vec{a_b}$$ $$ \vec{r_b}(0) = <0,h>$$ $$ \vec{V_b}(0) = <0,0>$$

A very good assumption for this experiment if it was performed at a small height (which is any height you could practically set up the experiment at), the velocity of the bullet will be much greater than the final velocity due to gravity. The convenient simplification that arises is:

$$|\vec{V_b}(t)| = (V_{bx}^2 + V_{by}^2)^{\frac{1}{2}} \approx V_{by}(t)$$

This allows a very nice closed form solution.

$$\vec{V_b}(t) = \left< \frac{V_p }{ C t V_p + 1 } , -g \frac{ \left( \frac{C t V_p }{2} + t \right) }{C t V_p + 1} \right>$$

From this, I can say the bullet will hit the ground at the same time as something dropped at the same time as firing when a certain condition is satisfied, which is:

$$ C t V_p \ll 1$$

Up until now I've used a non-standard drag coefficient. So I'll convert the equation to using the standard drag coefficient and then to ballastic coefficient.

$$ \frac{ C_d \rho A t V_p }{ 2 m} =\frac{ \rho t V_p }{ 2 BC} \ll 1 $$

So this is the first of my answers. For the other, I'll very simply address the angular sensitivity. The horizontal distance traveled by the bullet is much greater than the vertical distance it falls, but it still forms a triangle. We require that the displacement from the horizontal at the end of the flight due to the inclination of the gun must be much less than the height formally.

$$ V_p t \sin{\theta} \approx V_p t \theta \ll h$$

$$ \theta \ll \frac{h}{ V_p t} = \frac{1}{V_p} \sqrt{\frac{g h}{2}} $$

Say that the gun is at a height $h=20 m$. Bullet speed would fairly be $V_p=500 m/s$.

$$ \theta \ll 0.028 rad = 0.802^{\circ} $$

Note that this value is just a value sufficient to completely screw up the experiment so you would ideally want it 10x or more smaller than this. For the other part, dealing with the drag, I'll take some very general numbers. A drag coefficient for a generic cone is about $C_d = 0.5$, so I'll use that. For a bullet we'll say $m=10 g$. Air, $\rho=1.2 kg/m^3$. Bullet area we'll take as $A=0.2 m^2$. For 20 m height we have $t=2 s$.

$$\frac{ C_d \rho A t V_p }{ 2 m} = 0.006$$

So it seems my answer is that the gun must be held very still, but provided it's an "ordinary" gun and my numbers are correct, yes, the bullet will hit the ground at pretty much the same time. Hopefully I've formalized a little bit exactly when this is true.

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  • $\begingroup$ It would also be interesting to see if earth's curvature effects this. I'm guessing for handgun speeds it won't, but for high powered rifles it seems like it could $\endgroup$ – Oscar Smith Nov 11 '18 at 23:32
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To simplify this question and equalizing the atmospheric conditions, instead of bullets falling to the ground, consider yourself jumping up and falling back to the ground. Compare yourself (i) jumping on a stationary platform, and (ii) jumping on a fast moving train, do you think the two cases would be different? Do you feel lighter/heavier when you're on a train than when you're not? Why do you think so?

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I ran into this problem on Quora recently, and decided I would post an expanded version of the answer I gave there, here.

It is a common misconception that the vertical drag force only depends on the vertical velocity: but in the case of quadratic drag, this is not so. A simple diagram explains this:

enter image description here

I sketch here two objects, with same vertical velocity but different horizontal velocities - such that their total velocity is $v$ and $2v$ respectively. The drag force in both cases points along the direction of the velocity vector, but in one case it is 4x larger than in the other case. As a result, the vertical component of the drag force is different - even though the vertical velocity is the same!

The implication here is that when you fire a gun horizontally, the bullet will experience a significant increase in vertical drag force. You can compute the effect using a numerical integration - which I will show below.

The second effect to be considered is the rotation of the Earth. If you are standing on the equator, you appear to weigh less because part of the force of gravity is needed to keep you rotating with the earth; while you think you are standing still, you are actually moving at about 462 m/s to the East. If you shoot your gun Eastward, the bullet is traveling even faster - and it will appear "lighter". Shoot it to the West, and it will seem a little heavier. The effect is small. The acceleration due to the motion of the Earth is $\frac{v^2}{R_E}$, or about 0.033 m/s$^2$ which is about 1/300th of the acceleration due to gravity. If you shoot the bullet Eastward at 340 m/s, the speed of the bullet is added to the speed of rotation of the Earth - and that quadratic term starts to be noticed. In fact, the apparent acceleration towards the Earth $g$ will be 0.068 m/s$^2$ less than for the bullet that is dropped straight down: the flying bullet will appear "lighter". Since the time to fall scales with $\sqrt{\frac{2h}{g}}$, a change of 2 parts in 300 in gravity results in the bullet falling more slowly by about 1 part in 300. For a 1 meter drop, the time is roughly 0.44 seconds, and so the bullet shot East will drop more slowly by about 1.5 ms.

But that effect is much smaller than the effect of drag. For this example, I used the following parameters for the bullet (these are fairly realistic values for a larger pistol: rifles may have higher muzzle velocity, in which case the effect will be even bigger. These are adapted from http://gundata.org/blog/post/9mm-ballistics-chart/ ):

caliber:     9.0   mm
drag coeff:  0.149
mass:        7.5   g
velocity:  341     m/s

I wrote a little Python script that calculates the trajectory, taking into account the initial velocity and direction, drag, and the direction of the shot relative to the rotation of the earth. The result is the following plot:

enter image description here

There are five curves here. The ones marked "East" and "West" correspond to the effect of the rotation of the Earth, absent any drag. You can see the two bullets land almost at the same time; because there is no drag, the time taken is almost exactly the same as that for the one marked "Drop". The trajectory marked "Drag" shows how much longer it takes for the bullet to drop when it has horizontal velocity: this is the effect of quadratic drag, and results in the fired bullet taking significantly longer to drop. Zooming in on the interesting area makes it possible to see how small these effects are:

enter image description here

The difference between the bullet that is shot, and the one that is dropped, is 30 ms (bullet falls more slowly) without accounting for rotation of the Earth. Taking rotation into account, the different will get a little smaller or larger, depending on the location on Earth and the direction of the shot - with the biggest effect happening if you shoot East at the equator; that would add another 2 ms to the time taken to fall.

Finally, I computed the angle at which you would need to fire a bullet so it lands at the same time as the one dropped: this turns out to happen at an angle of about 0.05°. This corresponds to aiming the gun at a point that is 9 mm below the target at a distance of 10 m.$(^1)$

Doing this experiment with sufficient care to measure the effect is difficult, and obviously the above calculation made certain assumptions about the speed of the bullet, the mass, the drag coefficient etc. The fact that the Mythbusters measured 39.6 ms when the above calculation gave a value of 30 ms is very encouraging: I think that shows the drag effect is real.

(Incidentally, in my simulation the bullet would land at a distance of 125 m; the MB "drop zone" was at 110 m, but I don't know whether they were working with the same drop height: again, it's "really very close" though).

Source code of the simulation:

# example of numerical integration of projectile motion in 2D
# used to compute the difference in flying time for a bullet that is either
# shot or dropped
from math import sin, cos, atan2, pi, sqrt
import matplotlib.pyplot as plt

# function to compute drag:
def drag(v, theta):
    F =0.5*rho*v*v*A*Cd
    return (F*cos(theta), F*sin(theta))

# a function that does simple integration of the equation of motion:
def integrate(v, theta, m, x, y, g = 9.81, dt=0.01):
  vx = v * cos(theta)
  vy = v * sin(theta)   
  t = 0.

# storage for the result
  X = [x]
  Y = [y]
  T = [0.] 
  VX = [vx]
  VY = [vy]
  while ((y>0) | (vy>0)):
    # instantaneous force:
    Fx, Fy = drag(v, theta)
    # acceleration:
    ax = -Fx/m
    ay = -Fy/m - g
    # position update:
    x = x + vx*dt + 0.5*ax*dt*dt
    y = y + vy*dt + 0.5*ay*dt*dt
    # update velocity components:
    vx = vx + ax*dt
    vy = vy + ay*dt
    # new angle and velocity:
    v = sqrt(vx*vx+vy*vy)
    theta = atan2(vy,vx)
    # store result for plotting:
    X.append(x)
    Y.append(y)
    VX.append(vx)
    VY.append(vy)
    t = t + dt
    T.append(t)

# adjust last point to Y=0 - we may have "overshot":
  ft = Y[-2]/(Y[-2]-Y[-1]) # fractional time to last point
  X[-1] = X[-2] + (X[-1]-X[-2])*ft
  Y[-1] = 0.
  t = t - (1-ft)*dt
  T[-1]=t
  return (X,Y,T,VX,VY)


# *********
# run actual simulation

# fixed properties:
m = 0.0075     # mass
rho=1.22       # air density
g = 9.81       # gravitational acceleration
d = 0.009      # diameter of bullet
A = pi * d * d # cross sectional area
Cd = 0.149

# initial velocity & angle
v = 1120*0.3048 # 1120 fps
theta = 0.

# initial position, velocity, time
x = 0.
y = 1.    # height above target surface

# velocity of surface of earth at equator:
Re = 4.0e7/(2*pi)
Ve = 4.0e7/(24.*3600.) 

# acceleration due to shooting East, West:
a_east = -(Ve+v)*(Ve+v)/Re
a_west = -(Ve-v)*(Ve-v)/Re
a_still = -Ve*Ve/Re

Cd = 0.149

X,Y,T,VX,VY = integrate(0, theta, m, x, y, g+a_still) 

plt.figure()
plt.plot(T,Y)
print("T drop = %.1f ms"%(1000*T[-1]))

# look at the effect of shooting East, West without drag:
Cd=0.
X,Y,T,VX,VY = integrate(v, theta, m, x, y, g+a_east) 
plt.plot(T,Y)
print("T east = %.1f ms"%(1000*T[-1]))

X,Y,T,VX,VY = integrate(v, theta, m, x, y, g+a_west) 
plt.plot(T,Y)
print("T West = %.1f ms"%(1000*T[-1]))

Cd = 0.149
X,Y,T,VX,VY = integrate(v, theta, m, x, y, g+a_east) 
plt.plot(T,Y)
print("T drag = %.1f ms"%(1000*T[-1]))

for theta in [-0.0009]:
    X,Y,T,VX,VY = integrate(v, theta, m, x, y, g) 
    plt.plot(T,Y)
vx = VX[0]
vy = VY[0]
print("initial drag is %.2f N"%drag(sqrt(vx*vx+vy*vy),theta)[0])

print('Total flight time: %.3f sec\n'%T[-1])
print('Total distance: %.2f m'%X[-1])
print('Final horizontal velocity: %.2f m/s'%VX[-1])
print('Final vertical velocity: %.2f m/s'%VY[-1])
print('Final vertical drag: %.3f N'%drag(v, theta)[1])
print('Angle on impact: %.2f deg'%(180.*atan2(VY[-1],VX[-1])/pi))

plt.title('projectile motion')
plt.xlabel('Time (s)')
plt.ylabel('Height (m)')
plt.legend(('Drop', 'East','West','drag','aim down'))
plt.show()

(1)It is certainly possible to aim better than that - but it's not easy. For comparison, that's approximately the size of the "9" target area for the 10 m air rifle competition at the Olympics... to score a "10" you need to hit the dot in the middle, which means, with a 4.5 mm calibre, that you have a ± 4.5 mm tolerance. So it takes really top marksmanship, or a tripod, to aim the gun so accurately that this effect can be measured.

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You were told something which was not quite true, to illustrate a principle which is very true: In simple Newtonian motion, the equations of motion in one axis of a coordinate system do not have any effect on motion in any orthogonal axis.

Let me restate the case to make it just a bit more truthful. Lets say the bullet dropping to earth is motion in the 'z' direction. motion in the horizontal direcion could be 'x' or 'y' or a comination of the two. let's align our shooter so that he shoots exactly in the 'x' direction. We further define our coordinate system such that the 'x' direction remains locked in the direction of fire after the bullet is fired (does not turn or move with the Earth. Now lets place our shooter at the north (or south) pole, in a shooting range which is perfectly flat for at least 2000 feet. Our shooter is in a space suit, and the enclosed shooting range is pumped down to a vacuum. Finally, a mechanism is set up to drop a bullet at the exact time (t=0), and from the exact same height, as the shooter's bullet leaves the perfectly level gun barrel.

Both bullets will hit the range floor at exactly the same time (or so close you can't tell the difference).

The dropped bullet has a z component of velocity... and nothing else. It obeys the equation z(t) = z(0) - (a/2)(t^2); where z(0) = 6 feet, and 'a' = about 32 ft/(s^2). From a height of, say, 6 feet, it will hit the floor in about 0.25 seconds ("about 0.25..." --> I'm trying to keep the math simple).

The shot bullet has a z and an x component of velocity. The x component will be x(t) = (v)(t), where v = muzzle velocity = maybe 3000 ft/s. The motion in the 'z' direction is exactly yhe same as for the dropped bullet: z(t) = z(0) - (a/2)(t^2), and z(0) = 6 feet.

Now, here is the tricky part: when does the shot bullet hit the floor? That is motion in the z direction. Of the two equations governing motion of the bullet, only one has anything to do with vertical motion, and its soluiton is exactly the same as for the dropped bullet: It hits the floor in about 0.25 seconds. In that amount of time it will have traveled 750 feet in the x direction and will still have a forward velocity of 3000 ft/s.

The truth that a shot bullet and a dropped bullet fall at the same rate (neglecting secondary effects) is an important phenomenon, which must be taken into account for long range firing. If a rifle is "sighted in" for a 100 yard target, it will shoot noticably lower (roughly 4 inches lower for a medium power rifle) on a 200 yard target.

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The bullets time in the air, having been fired horizontally, depends on it's velocity. In general that varies from about 900 fps to near 3000 fps for a rifle, so obviously the time it takes to hit the ground varies as well.

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  • $\begingroup$ fps? Frame per second? I would highly recommend to use SI-unit (meter/second), or maybe derived km/hour. $\endgroup$ – Bernhard Jul 13 '14 at 8:09
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    $\begingroup$ @Bernhard well I'd say what's important is not to use SI units specifically, but to be clear about what units are being used. fps is presumably feet per second (which is common for bullet speeds). $\endgroup$ – David Z Jul 13 '14 at 14:45

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