Fired and dropped

There has been a great deal of discussion about this problem in the solutions. Many people have invoked an experiment performed by the Mythbusters several years back which showed that a fired bullet and a dropped bullet hit the ground at the same time. We here at Brilliant love the Mythbusters, but we love physics more. Let's break this problem down a bit:

Quadratic Drag

As some of the solutions have indicated, the interesting bit of physics in this problem comes from the fact that the drag force from air resistance is proportional to the squared velocity of the bullet. The proportionality constant depends on the object's geometry, but for our purposes we'll just call the drag coefficient $k$ . If we break down the velocity into its $v_x$ and $v_y$ components, that means the drag term is something like this at some moment after the bullet has left the barrel.

$F_d = -k|v|^2,$

where $|v| = \sqrt{v_x^2 + v_y^2}$ .

Breaking down the force

Like any other force, we can break down the drag force into components along the $x$ and $y$ axes. If the bullet's trajectory is off from horizontal by an angle $\theta$ , then the $y$ component of the force is $F_d \sin{\theta}$ and the $x$ component of the force is $F_d \cos{\theta}.$ To avoid any unnecessary variables, we can express these trigonometric functions using the bullet's $x$ and $y$ components. Using the right triangle above, we can observe that

$\sin{\theta} = \frac{v_y}{|v|}, \\ \cos{\theta} = \frac{v_x}{|v|}.$

When substituted into the $x$ and $y$ components of the drag force, the divided factor of $|v|$ simplifies and yields $x$ and $y$ components of $-k |v| v_x$ and $-k |v| v_y$ respectively. Even though these components of the force are orthogonal, notice that they both still depend on the total velocity $|v|$ in addition to their velocity component.

Equations of motion

Now that we've broken down the drag force into the $x$ and $y$ components, we can construct the equations of motion according to Newton's second law:

$\begin{aligned} m \, v_x'(t) &= -k |v| v_x \\ m \, v_y'(t) &= -mg - k |v| v_y \end{aligned}$

Substituting $|v|$ as the norm of the two spatial components:

$\begin{aligned} m \, v_x'(t) &= -k v_x \sqrt{v_x^2 + v_y^2} \\ m \, v_y'(t) &= -mg - k v_y \sqrt{v_x^2 + v_y^2} \end{aligned}$

You might notice now that these equations are coupled through $|v|$ , and likely can't be solved explicitly. Even without solving, we can consider what effect the total velocity of the bullet $|v| = \sqrt{v_x^2 + v_y^2}$ contributes to the drag term in both components of the bullet's velocity. If the bullet is fired from a gun, a large horizontal velocity will increase the vertical drag force that resists the pull of gravity. If a bullet is simply dropped, the horizontal velocity will be zero, and the total velocity will be much lower. This would yield a weaker drag force to resist gravity, and explains why a dropped bullet hits the ground quicker

Finding the trajectories

Though explicitly solving these coupled differential equations is nasty, we can use a numerical solver in Mathematica (or your computational tool of choice, my notebook that generates these plots and zeros is here ). Picking some physical constants for the drag coefficient $k$ , the mass $m$ and an initial height yields the following trajectories for a dropped bullet and one fired at $\SI{500}{\meter \per \second}$ .

Fired bullet trajectory and time dependence of height (Quadratic drag)

Dropped bullet trajectory and time dependence of height (Quadratic drag)

The fired bullet hits the ground after about 0.60 seconds, while the dropped one hits after only 0.45 seconds.

What about linear drag?

If we were to use a simpler linear approximation of air drag, the total velocity would no longer factor in to the drag of both components, and the equations would no longer be coupled. I leave that to the reader to construct similar equations of motion using a linear drag term like $F_d = -k |v|$ . The trajectory of such a bullet looks something like the one below, and hits the ground just like the dropped one, after about 0.45 seconds.

Fired bullet trajectory and time dependence of height (Linear drag)

This 0.45 seconds figure is about the same as the time for a bullet to drop from a height of 1 meter neglecting all air drag: ( $\frac{1}{2} g (\SI{0.45}{\second})^2 = \SI{1}{\meter}$ ). This kind of calculation exemplifies why in many cases where the velocity of an object is low, air drag can be neglected without much loss of accuracy. But when a bullet is fired at hundreds of meters per second, the contribution of quadratic drag becomes significant.

By a slight abuse of notation, let $v = (v_x,v_y)$ denote the velocity of either given bullet at any given time.

We know that the drag acts against the velocity and is proportional to $|v|^2.$ Therefore the vertical component of the drag is proportional to $|v|v_y.$ For the dropped bullet, this equals $v_y^2$ but for the fired bullet, it will equal $v_y^2\sqrt{1+\left(\frac{v_x}{v_y}\right)^2} > v_y^2$ .

That is, for a given vertical component of the velocity, the vertical component of the drag would be greater for the fired bullet than for the dropped bullet, so the resulting downward acceleration for the fired bullet will be less than the downward acceleration for the dropped bullet.

Therefore, the dropped bullet will hit the flat plane first.

Dropped bullet falls faster as the drag resistance for it is less than for the fired bullet.

The falling bullet gets help from gravity, but the shotted one has only its starting velocity, that decreases over time. Thus falling one speeds up, to some point. However I think, there might be some limit short distance, in witch the shotted bullet would be faster than falling, because falling one would not speed up fast enough.

Let us take the positive $y$ direction downwards. Assuming turbulent flow, the air resistance acting on the dropped bullet is $F=-\alpha v^2$ , where $\alpha$ depends on the density of the air and the shape of the bullet and $v_y$ is the velocity of the bullet. The velocity is downward and the force is upward. For the fired bullet the air resistance is similar, $F=\alpha v_{tot}^2=\alpha (v_x^2+v_y^2)$ . The vertical component of this force is $F_y=-\frac{v_y}{v_{tot}}\alpha v_{tot}^2=-\alpha v_y \sqrt{v_x^2+v_y^2}$ , where the negative sign represents the fact that $v_y$ is downward and $F_y$ is upward.

In both cases the drag force reduces the acceleration due to gravity, but for the fired bullet the reduction is larger by a factor of $v_{tot}/v_y$ . Accordingly, if we make graphs of $v_y$ vs. $t$ , the line for the dropped bullet falls a little bit below the straight line that corresponds to the frictionless case, $v_y=gt$ (see red curve below). On the graph for the fired bullet, at any value of $v_y$ the slope must be smaller than for the dropped bullet, therefore this line will be below the line corresponding to the dropped bullet (see blue curve below). Since the area under the curves corresponds to the distance traveled, the dropped bullet covers more distance in the same time, and therefore reaches the plane first.

Comments:

1. As some of the commenters pointed out, there is a Mythbuster episode that addresses this exact same question. They argue that the two bullets arrive to the ground at the same time. If you watch carefully, at the end they say that there was a time difference of 39ms. Computer simulations show that the expected time difference is about the same. While this time difference looks small, it is not zero, especially if it is measured relative to the time it takes for the bullets to drop, 440ms. The difference is 10%. There is also a numerical simulation , where the time difference is calculated, and the result is 0.01s. The difference between the Mythbusters and the simulation is probably due to an error in the horizontal positioning of the gun.

2. How large is the drag force relative to the weight of the bullet? The drag force is $F=c_D \rho v^2 A$ , where $c_D$ is the drag coefficient, $\rho$ is the density of air, $v$ is the velocity and $A$ is the area facing the airflow. We take a bullet that is a 1cm diameter sphere, and select a velocity of $v=300m/s$ , which is typical of a handgun, slightly less than the speed of sound. For a sphere $c_D=0.47$ and the density of air is $\rho=1.25kg/m^3$ . The drag force is 4.15N. In contrast, the weight is $W=\rho_{lead}V g$ , where $\rho_{lead}=11.3 \times 10^3 kg/m^3$ is the density of the bullet, $g$ is the acceleration of gravity and $V$ is its volume. We get $W=0.058N$ , that is 70 times less than the drag force. Clearly, the drag force is a very important part of this problem.

3. Mark Lyttle asked: How do we know that the air flow is turbulent for the dropped bullet? That is a very good question. The answer depends on the Reynolds number, defined as $Re=v d/\nu$ , where $v$ is the velocity, $d$ is the diameter of the bullet and $\nu=1.5\times 10^{-5}m^2/s$ is the kinematic viscosity of the air. For Reynolds numbers larger than about 100 the flow is turbulent, and the drag force is $F=c_D \rho v^2 A$ , where $c_D$ is a coefficient that depends on the shape of the object ( $c_D=0.47$ for a sphere), $A$ is the area facing the flow and $\rho$ is the density of air . For low Reynolds numbers the flow is laminar, the drag force is proportional to the velocity and it is always less than the drag force calculated for turbulent flow.
   Assuming we drop the bullet from a height of 2m, the velocity at the ground will be 6.3m/s. With $d=1cm$ the Reynolds number is 4,200 and the flow is turbulent. There is no question that the flow is laminar at the very beginning of the motion. That means that the drag force is less than the force we assumed and the dropped bullet reaches the ground in a little bit less time. Our conclusion is still valid.
   

At any instance in time, the drag force, d, is proportional to the square of the instantaneous speed, v, and acting in the opposite direction to the instantaneous direction of travel. $d = -k v^2 \ \ \ (1)$ where k is an arbitrary scalar constant and d and v are scalar. But since we have mentioned direction, we recognise that the bullet has a velocity vector $V$ , and the drag has a force vector $D$ .

  D    Dv
   \   |
    \d |
     \ |
alpha(\|
Dh ----*----Vh
       |\)theta
       | \
       |  \v
       |   \
       Vv   V

apologies for the ASCII art; I have not worked out how to insert a diagram

We can break the vector velocity, $V$ , of the bullet with speed $v$ (magnitude) and at an angle $\theta$ below horizontal (direction) into its vertical (downward) and horizontal components, $V_v$ and $V_h$ respectively, where $v = \sqrt{V_h^2 + V_v^2} \ \ \ (2)$ $\theta = \sin^{-1}(\frac{V_v}{v}) \ \ \ (3)$

Similarly we can break the drag force vector, $D$ , with magnitude d and at an angle $\alpha$ above horizontal (direction), into its vertical (upward) and horizontal components (back towards the gun), $D_v$ and $D_h$ respectively, where $\sin(\alpha) = \frac{D_v}{d} \ \ \ (4)$ and we are only interested in the vertical component, $D_v$ , because it is only that which determines the time it takes for the bullet to reach the plane. The drag force $D$ acts in exactly the opposite direction to the velocity vector $V$ , so $\alpha = \theta \ \ \ (5)$

The following steps are purely algebra.

From (4) and (5) $D_v = d \sin(\theta) \ \ \ (6)$ From (1) and (6) $D_v = -k v^2 \sin(\theta) \ \ \ (7)$ From (3) and (7) $D_v = -k v^2 \sin(\sin^{-1}(\frac{V_v}{v})) \ \ \ (8)$ Since $\sin(\sin^{-1}(x)) = x$ $D_v = -k v^2 (\frac{V_v}{v}) \ \ \ (9)$ From (2) and (9) $D_v = -k (V_h^2 + V_v^2) (\frac{V_v}{\sqrt{V_h^2 + V_v^2}}) \ \ \ (10)$ Since $\frac{x}{\sqrt{x}} = \sqrt{x}$ $D_v = -k V_v \sqrt{V_h^2 + V_v^2} \ \ \ (11)$ When the horizontal component $V_h = 0$ , the case where the bullet is dropped, this reduces to $D_v = -k V_v^2 \ \ \ (12)$

Comparing (11) and (12) we see that the vertical component of the drag force, $D_v$ , contains an additional element $V_h$ in (11), for the case for the bullet which is shot from a gun.

Since the vertical component of the drag force is larger, then the bullet shot from the gun will take longer to reach the ground, hence the dropped bullet reaches the plane first.

In words: Since the drag force depends on the square of the speed, comprising both horizontal and vertical components, then the horizontal component of the speed of the bullet cannot be eliminated from the vertical drag force by the normal rules for separating horizontal and vertical components of the velocity and force vectors.

This discussion is very similar that presented by Blake Farrow, but the conclusion is reached without using the |v| notation, and without resorting to the equations of motion.

This problem was a challenge to me. The correct solution appears counter intuitive and I had thought that Brilliant had got it wrong. I would have been bitterly disappointed with Brilliant if that was the case, so I went about trying to prove that Brilliant had made a mistake. I ended up concluding that Brilliant were indeed correct.

The answer to this question does depend on whether the air flow against the bullets is laminar or turbulent. First we make a realistic assumption that the bullet speeds are high and flow is turbulent. Next, the most important concept required is that the time of flight for each bullet is completely determined by the net acceleration in the vertical direction. If there were no air resistance, then irrespective of how fast the fired bullet is in the horizontal direction, both bullets would have reached the ground at the same time. Now the force of air resistance is proportional to the square of the magnitude of the velocity of the bullet. The fired bullet has an additional x-component of velocity, which increases its magnitude of the velocity vector compared to the dropped bullet. In case of the dropped bullet, the drag force is along the vertical direction and is = $K_{\rm T}.v_{\rm y}^2$ , where $K_{\rm T}$ is the proportionality constant and $v_{\rm y}$ is the magnitude of velocity along the y-direction. In case of the fired bullet, the drag force is a vector directed opposite to the net velocity vector. Therefore, we must extract the y-component of this force. The y-component of drag is = $K_{\rm T}.v_{\rm y}.{(v_{\rm y}^2+v_{\rm x}^2)}^{1/2}$ . Clearly, the y-component of the drag is always higher for the fired bullet than that for the dropped one. Hence, the net downward acceleration is less for the fired bullet, which means it will take a longer time to reach the ground. So, the dropped bullet reaches ground first. Now if we assume that bullet speeds are low such that air flow is laminar, then we can derive that the y-component of the drag forces in both bullets are the same: $K_{\rm L}.v_{\rm y}$ . Thus, in such a scenario both bullets will reach the ground at the same time. In a qualitative way one can visualize this outcome as follows: The air drag will create discrepancy in the time of flight for the two bullets only when their speeds reach high enough value for the air flow to enter turbulent regime. In real life this means: increasing the height of launch or increasing the muzzle speed of the gun.

If the fired bullet had sufficient velocity, it would never hit the ground because of curvature of earth. If the ‘fired’ bullet had zero velocity, it would hit the ground at the same time as dropped bullet. As a ‘fired’ bullet must have SOME velocity, the dropped one must always hit the ground first.

If the bullet is fired parallel to the horizontal plane it will take a lot longer to hit the plane. The bullet has to lose velosity, slow down, and cover more distance ,then fall onto plane, where the dropped bullet has a shorter path to the horizontal plane.

as a person with enough speed can remain on water speed can remain everything in the air

The dropped bullet drops first because the fired bullet travels further horizontally through the barrel. Also, shooters will deflect their arms slightly up from the force of the powder ignition.

I figured both bullets have equal gravitational forces working on them but only the fired bullet has the inertia from the gun fire forcing it in the direction the gun was pointing. Since the Earth is round, the fired bullet can't truly project parallel to the ground and is actually gaining altitude for a time while the unfired bullet does not

I was going to say, " whichever one will hit my groin at the end.

The drag resistance from the dropped bullet makes it fall faster, While the shotted bullet only gets help from gravity.

This might not be correct thinking but it got me the right answer.

Because the dropped bullet will be free to orient itself efficiently as it falls, or with tumble, it will on average, have a more efficient wind profile in the "Up-Down" direction than will the bullet that was fired from a gun.

The bullet which was fired from the gun will be oriented horizontal to the flat plane at all times. That profile is good for vertical travel, but not as good in the for horizontal travel.

In a vacuum, I believe these would hit the plane at the same time since the bullet's orientation as it falls won't matter.

Fired bullet have some energy which oppose gravity while dropped bullet don’t have.

As it defys sound barrier its most likely to continue on as gravity cannot pull it immediately

This solution is using a couple of rules and Concepts of gun ballistics and Newton's laws of inertia.:

1: a gun has rifling to increase the spin, this decreasing the drag of air resistance of a bullet after it leaves the barrel. A link here: https://www.google.com/amp/s/io9.gizmodo.com/5944455/the-physics-of-bullets-why-a-modern-gun-shoots-10-times-further-than-its-19th-century-counterpart/amp more accurately explains the physics than I can; however, to sum it up, the bullet acts much like a spinning top going sideways through the air, stabilizing it and keeping it rotating on a horizontal axis. This causes effects like the Eötvös effect and the Coriolis effect. The former effect causes objects and mediums not secured to the ground to be more affected by gravity when moving to the West or less to the east. The latter, or the Coriolis effect, cause a bullet to spin left or right depending on the rotation of the rifling of the gun. Fortunately, or unfortunately the Coriolis effect has no effect in this problem.

2: the bullet traveling through the air has more horizontal velocity at the beginning, thus is less affected by gravity on its mass until it overcomes it's horizontal vertical Velocity in relation to its initial vertical inertia, using Newton's second law in relation to rotational inertia. A much easier explanation: the bullet fired from the gun has more horizontal Force thus increasing the distance it has to travel before falling to the ground; while the bullet released from the hand has a straight vertical path to the ground.

it is easy, the bullet dropped has the less distance to cover until it hits the ground, while the fired bullet has more velocity than the dropped one causing it to go farther, thus taking more time to land.