Effect of drag on projectile flight time - simple solution?

Question

I saw a question like this in an AS-level physics paper (consisting of 40 questions given to ~17-year-olds, with a 75-minute time limit):

Two identical balls, x and y, are thrown up from the earth's surface with the same velocity. x is thrown in the air and y is in a vacuum. Which ball reaches a higher maximum height and which ball will hit the ground first?

It's immediately obvious that drag makes x reach the (lower) peak faster, and generally, you'd think that a shorter descent would take less time. But drag will also make x's acceleration slower than that of y's during their descents.

How could you go about proving (quickly,) if drag causes a net increase or decrease in time?

Answer 1

I don't think this can be easily proven - I think it depends on the viscosity of air and the speed with which the projectile is fired.

Under ordinary circumstances, I'm sure the main factor is that air resistance takes energy out of the system early on when the ball is moving fast, and this makes it come down sooner. Imagine hitting a badminton shuttlecock with a golf club - it's not going to stay in the air for anywhere near as long as a golf ball can, even if they have the same mass and initial velocity.

But consider shooting a bullet upwards into a vat of treacle. It's just going to get lodged in the middle of the syrup and only make its way back down again very slowly, over hours or days, taking much longer than it would if the treacle weren't there.

Now treacle probably differs from air not only in the coefficient of drag but also in the scaling law for how the drag force depends on velocity, so it might not be a fair comparison. But still, I'd be willing to bet that even if you use the same equation and just change the drag coefficient, you'll find that which ball lands first depends on the parameters. If so then there will not be an easy proof either way.

In other words, unless I'm missing something, it's a bad question.

Answer 2

The thick red line corresponds to the velocity of the ball without air resistance. Its gradient is $-g$.

It should be clear that with air resistance, you decelerate faster. Therefore you end up with something like the thin red line. Obviously, the thin red line hits the peak (where ($v = 0$)) faster (for smaller $t$).

The distance travelled is the area under the curve. Without air resistance, the time taken to go up equals the time taken to go down. It should, therefore, take the thin black line to return to the ground.

With air resistance, we need the blue zone to have the same area as the green zone (where the right-hand boundary of the green zone can be extended arbitrarily), and it should be evident that this will happen well before the boundary reaches the thin black line. One could imagine the air resistance being a lot larger, but it's hard to draw an appropriate thin red line such that the gradient where it crosses the $v=0$ axis is still approximately $-g$.

This isn't a rigorous proof, of course, just an intuitive one.

Answer 3

Two identical balls, x and y, are thrown up from the earth's surface with the same velocity. x is thrown in the air and y is in a vacuum. Which ball reaches a higher maximum height and which ball will hit the ground first?

How could you go about proving (quickly, and with the knowledge of a 17 year old,) that drag causes a net increase in time

You can use the graphical method suggested in one of the other answers, or you could choose a specific simple form for the drag (e.g., linear) and solve the equation: $$ \dot v = -g -\alpha v\;,\tag{1} $$ where $\alpha$ is a small positive constant (i.e., $\alpha v/g << 1$), because we assume the drag force due to air is small.

Integrating Eq. (1) twice provides an exact equation for the position as a function of time. If we take the position of "the ground" (a.k.a. "earth's surface") as $y=0$, and if we take the starting time as $t=0$, this equation can be solved for the times that $y=0$. One such time is $t=0$. The other time when $y=0$ generally would have to be solved for numerically, since the equation for the position is transcendental: $$ y(t) = -\frac{gt}{\alpha} - (\alpha v_0 + g)\frac{1}{\alpha^2}\left(e^{-\alpha t} - 1\right)\;. $$

But, by Taylor expanding about $\alpha=0$, the other solution can be obtained to lowest order in $\alpha$ by just using some algebra: $$ t_G = \frac{2 v_0}{g} - \alpha \frac{2v_0^2}{3g^2}+\mathcal{O}(\alpha^2)\; $$ Thus, when the drag force is small, the ball in air hits the ground slightly sooner than the ball in vacuum. (Note, although this result was derived for infinitesimal linear drag, one of the other answers has asserted that any infinitesimal power-law drag will lead to the ball in air hitting the ground slightly sooner.)

Answer 4

There is no simple intuitive explanation, because it's not true!

As explained in this paper, the answer depends on the model of air drag. If the drag force is proportional to $v^n$, it turns out the trajectory with drag always takes less time for $n \geq 1$, but for $n < 1$ it depends on the initial speed. (This makes intuitive sense, as when $n$ is high, the drag force rises quickly with speed. The speed will tend to be higher on the way up than the way down, so the effect of the drag force is more important on the upward part, where it points down.)

Answer 5

I don't think one can prove "that drag causes a net increase in time" without some additional assumptions.

Let us consider the following example. Let us choose the initial speed of both balls barely exceeding the escape speed. Then the ball moving in vacuum will never fall on Earth, unlike the ball moving in air.

Answer 6

TL;DR: In this answer we consider an analytic approach beyond infinitesimal drag. OP's competition between damped fall vs. free fall turns out to be rather subtle.

1. Consider the 1D vertical damped motion$^1$ $$ \dot{v}~=~-g-f(v), \qquad\dot{x}~=~v,\tag{1}$$ where $f(v)=-f(-v)$ is a drag profile, e.g. linear or quadratic drag.

2. In the spirit of this Phys.SE post, we can find a Hamiltonian formulation of eqs. (1). The Hamiltonian is $$ H(x,v)~=~T(v)+V(x),\tag{2}$$ where $$ V(x)~:=~x, \qquad T(v)~:=~\int_0^v\!\frac{\tilde{v}d\tilde{v}}{g+f(\tilde{v})},\tag{3}$$ with symplectic potential $$ \vartheta(v)~:=~\int_0^v\!\frac{d\tilde{v}}{g+f(\tilde{v})},\tag{4}$$ and fundamental Poisson bracket $$ \{x,v\}~=~g+f(v)~=~\frac{1}{\vartheta^{\prime}(v)}. \tag{5}$$ The Hamiltonian Lagrangian is $$ L_H(x,\dot{x},v)~=~\vartheta(v)\dot{x}-H(x,v).\tag{6}$$ It is straightforward to check that the corresponding Euler-Lagrange (EL) equations/Hamilton's equations are precisely eqs. (1).

3. Let there be given boundary conditions $x_i=0=x_f$ and initial upward velocity $v_i>0$ of the upward throw. We can find the final downward velocity $v_f<0$ from the fact that the Hamiltonian (2) is a constant of motion: $$ T(v_i)~=~T(v_f). \tag{7}$$ The time of flight $\Delta t$ is given by the symplectic potential $$\begin{align}\Delta t~=~&\int_{v_i}^{v_f}\!\frac{dv}{\dot{v}}\cr ~\stackrel{(1)}{=}~&\int_{v_f}^{v_i}\!\frac{dv}{g+f(v)} ~\stackrel{(4)}{=}~\vartheta(v_i)-\vartheta(v_f).\end{align}\tag{8}$$ Alternatively, if it is difficult to solve eq. (7), one can try to solve for the time of flight $\Delta t$ via this equation $$\begin{align} 0~=~&\Delta x~=~\int_{t_i}^{t_f} \!dt~v(t)\cr ~\stackrel{(8)}{=}~&\int_0^{\Delta t}\!d\widetilde{\Delta t}~\vartheta^{-1}\left(\vartheta(v_i)-\widetilde{\Delta t}\right).\end{align}\tag{9}$$

4. Example: Constant drag $f(v)=b{\rm sgn}(v)$ where $b<g$: Kinetic term $$ T(v)~\stackrel{(3)}{=}~\frac{v^2}{2(g+b{\rm sgn}(v))}.\tag{10}$$ Final velocity: $$ v_f~\stackrel{(7)}{=}~-v_i\sqrt{\frac{g-b}{g+b}}.\tag{11}$$ Symplectic potential: $$ \vartheta(v)~\stackrel{(4)}{=}~\frac{v}{g+b{\rm sgn}(v)}.\tag{12}$$ Time of flight: $$ \Delta t~\stackrel{(8)}{=}~\frac{v_i}{\sqrt{g+b}}\left(\frac{1}{\sqrt{g+b}}+\frac{1}{\sqrt{g-b}}\right).\tag{13}$$ It is easy to check that the damped fall is faster than the free fall iff $0<b<\frac{g}{\sqrt{2}}$, cf. Ref. 1.

5. Example: Quadratic drag $f(v)=b|v|v$: Kinetic term $$ T(v)~\stackrel{(3)}{=}~\frac{{\rm sgn}(v)}{2b}\ln\left|1+\frac{b}{g}|v|v\right|.\tag{14}$$ Final velocity: $$ \sqrt{\frac{b}{g}}|v_f|~\stackrel{(7)}{=}~\sqrt{1-\frac{1}{1+\frac{b}{g}v_i^2}}.\tag{15}$$ Symplectic potential: $$ \vartheta(v)~\stackrel{(4)}{=}~\left\{\begin{array}{rcl}\frac{1}{\sqrt{gb}}\arctan\left(\sqrt{\frac{b}{g}}v\right)&\text{for}&v~\geq~0,\cr -\frac{1}{\sqrt{gb}}{\rm artanh}\left(\sqrt{\frac{b}{g}}|v|\right) &\text{for}&v~\leq~0.\end{array}\right.\tag{16}$$ Time of flight: $$\begin{align} \Delta t~\stackrel{(8)}{=}~&\frac{1}{\sqrt{gb}}\arctan\left(\sqrt{\frac{b}{g}}v_i\right)\cr &+\frac{1}{\sqrt{gb}}{\rm artanh}\left(\sqrt{1-\frac{1}{1+\frac{b}{g}v_i^2}}\right).\end{align}\tag{17}$$ It is easy to check that the damped fall is always faster than the free fall.

6. Example: Linear drag $f(v)=bv$: Kinetic term $$ T(v)~\stackrel{(3)}{=}~\frac{v}{b}-\frac{g}{b^2}\ln|1+\frac{b}{g}v|.\tag{18}$$ The eq. (7) for the final velocity $v_f$ is transcendental. Symplectic potential: $$ \vartheta(v)~\stackrel{(4)}{=}~\frac{1}{b}\ln|1+\frac{b}{g}v|.\tag{19}$$ Eq. (9) for the time of flight $\Delta t$ becomes $$ \Delta t~=~(\frac{b}{g}v_i+1)\frac{1-e^{-b\Delta t}}{b}.\tag{20}$$ Analysis of eq. (20) shows that the damped fall is faster than the free fall, cf. answer by AccidentalTaylorExpansion.

7. Example: Infinitesimal power drag $f(v)=b|v|^n{\rm sgn}(v)$ where $b$ infinitesimal: Kinetic term $$ T(v)~\stackrel{(3)}{=}~\frac{v^2}{2g}-\frac{b{\rm sgn}(v)}{g^2}\frac{|v|^{n+2})}{n+2}+{\cal O}(b^2).\tag{21}$$ Final velocity: $$ v_f~\stackrel{(7)}{=}~-v_i\left(1-\frac{2b}{g}\frac{v_i^n}{n+2}+{\cal O}(b^2)\right).\tag{22}$$ Symplectic potential: $$ \vartheta(v)~\stackrel{(4)}{=}~\frac{v}{g}-\frac{b}{g^2}\frac{|v|^{n+1}}{n+1}+{\cal O}(b^2).\tag{23}$$ Time of flight: $$ \Delta t~\stackrel{(8)}{=}~\frac{2v_i}{g}-\frac{2b}{g^2}\frac{v_i^{n+1}}{n+2}+{\cal O}(b^2).\tag{24}$$ For infinitesimal drag, the damped fall is always faster than the free fall.

References:

1. C.E. Mungan, S.T. Rittenhouse & T.C. Lipscombe, Am. J. Phys. 89 (2021) 67-71; App. B. It would have been interesting if the authors had included a third $b$-axis in their Fig. 3. $\qquad$ (Hat tip: knzhou.)

---

$^1$ Here we have for simplicity put mass $m=1$. To re-instate mass-dependence, replace $f\to \frac{f}{m}$ or $b\to \frac{b}{m}$.

Answer 7

I'll start with the aside that the question neglects buoyancy; if the "ball" is a party balloon, it could remain aloft for a long time, or indefinitely if it is lighter than air.

Buoyancy aside, my intuition was similar to N. Virgo's, that with sufficiently high drag and initial velocity, it should be possible to get the ball-in-air case to "punch" through to a high altitude, from which it would then descend slowly at terminal velocity, slowly enough for the ball-in-vacuum case to catch up and win.

But experimentally, that is wrong. I wrote a simple simulator in Python, and played with the parameters, and found that as long as drag is at least linear, the ball-in-air always wins. Adding superlinear drag only makes the ball-in-air return to the ground even faster due to quickly arresting the ascent, after which the linear term dominates during terminal velocity descent. Adding more initial velocity merely increases the advantage for the air case.

But here's the weird thing, which makes me think there may be a simple analytical argument hiding here: regardless of the parameters, the velocity of the balls is the same at the moment the first one hits the ground! That is, the ball-in-air consistently hits the ground, at its terminal velocity, just as the ball-in-vacuum is reaching that same speed, as it begins its descent!

Example 1:

inputs:
  initialVelocity    :       100.00 m/s
  gravity            :       -10.00 m/s^2
  timeStep           :         0.10 s
  linearResistance   :         0.10 1/s
  quadraticResistance:         0.00 1/m

outputs:                            Annotated comments
  winner: B (air)                   -------------------
  t     :        16.00 s            Time when B hits the ground.
  xA    :       328.00 m            Height of A.
  max_xA:       505.00 m            Maximum height of A.
  vA    :       -60.00 m/s  <---    Velocity of A.
  vAturn:        10.10 s            When A turned around.
  xB    :        -0.55 m            Height of B (it hit the ground).
  max_xB:       310.33 m            Maximum height of B.
  vB    :       -59.94 m/s  <---    Velocity of B (~terminal).
  vBturn:         6.90 s            When B turned.

Notice that the final vA and vB are nearly the same. A (in vacuum) of course would keep accelerating downward if the simulator continued, while B has long since reached its terminal velocity.

Example 2 (10x faster initial velocity):

inputs:
  initialVelocity    :      1000.00 m/s
  gravity            :       -10.00 m/s^2
  timeStep           :         0.10 s
  linearResistance   :         0.10 1/s
  quadraticResistance:         0.00 1/m

outputs:
  winner: B (air)
  t     :       110.00 s
  xA    :     49555.00 m
  max_xA:     50050.00 m
  vA    :      -100.00 m/s  <---
  vAturn:       100.10 s
  xB    :        -0.17 m
  max_xB:      7614.13 m
  vB    :       -99.98 m/s  <---
  vBturn:        23.90 s

Again, the final vA and vB are nearly the same.

Example 3 (2x linear drag coefficient):

inputs:
  initialVelocity    :       100.00 m/s
  gravity            :       -10.00 m/s^2
  timeStep           :         0.10 s
  linearResistance   :         0.20 1/s
  quadraticResistance:         0.00 1/m

outputs:
  winner: B (air)
  t     :        14.20 s
  xA    :       418.90 m
  max_xA:       505.00 m
  vA    :       -42.00 m/s     <---
  vAturn:        10.10 s
  xB    :        -2.58 m
  max_xB:       228.11 m
  vB    :       -41.48 m/s     <---
  vBturn:         5.50 s

I suspect this is true in general for linear resistance, which would then prove that the ball-in-air always lands first. But I don't know how to prove this, especially using high school physics math.

Simulator code:

#!/usr/bin/env python3
"""Simulate two thrown balls."""


# Initial upward velocity in m/s.
initialVelocity = 100

# Uniform gravity in m/s^2.
gravity = -10

# Simulation time step in s.
timeStep = 0.1

# Linear air resistance factor in 1/s.  That is, ball B experiences
# acceleration of `-vB*linearResistance`.
linearResistance = 0.1

# Quadratic air resistance factor in 1/m.  B experiences acceleration
# of `-abs(vB)*vB*quadraticResistance`.
quadraticResistance = 0

# True to print every time step.
debug = False


def simLoop():
  # Elapsed time.
  t = 0

  # A is in a vacuum.
  xA = 0
  vA = initialVelocity
  max_xA = 0
  vA_turned_time = None

  # B is in air.
  xB = 0
  vB = initialVelocity
  max_xB = 0
  vB_turned_time = None

  # Who hits first?
  winner = None

  if debug:
    print(f"              [-------- vacuum --------]  [--------- air ----------]")
    print(f"           t            xA            vA            xB            vB")
    print(f"------------  ------------  ------------  ------------  ------------")

  while winner is None:
    t += timeStep

    # Update A by one time step.
    xA += vA * timeStep
    vA += gravity * timeStep
    if xA > max_xA:
      max_xA = xA
    if vA_turned_time is None and vA < 0:
      vA_turned_time = t

    # Update B by one time step.
    xB += vB * timeStep
    vB += (gravity
             - vB * linearResistance
             - abs(vB) * vB * quadraticResistance) * timeStep
    if xB > max_xB:
      max_xB = xB
    if vB_turned_time is None and vB < 0:
      vB_turned_time = t

    # Current system state.
    if debug:
      print(f"{t:12.2f}  {xA:12.2f}  {vA:12.2f}  {xB:12.2f}  {vB:12.2f}")

    # Winner?
    if xA <= 0:
      if xB <= 0:
        winner = "Tie"
      else:
        winner = "A (vacuum)"
    elif xB <= 0:
      winner = "B (air)"

  if debug:
    print("")

  vA_rising = "  A is still rising!" if vA > 0 else ""
  vB_rising = "  B is still rising!" if vB > 0 else ""

  print("inputs:")
  print(f"  initialVelocity    : {initialVelocity:12.2f} m/s")
  print(f"  gravity            : {gravity:12.2f} m/s^2")
  print(f"  timeStep           : {timeStep:12.2f} s")
  print(f"  linearResistance   : {linearResistance:12.2f} 1/s")
  print(f"  quadraticResistance: {quadraticResistance:12.2f} 1/m")
  print("")

  print("outputs:")
  print(f"  winner: {winner}")
  print(f"  t     : {t:12.2f} s")
  print(f"  xA    : {xA:12.2f} m")
  print(f"  max_xA: {max_xA:12.2f} m")
  print(f"  vA    : {vA:12.2f} m/s{vA_rising}")
  if vA_turned_time is not None:
    print(f"  vAturn: {vA_turned_time:12.2f} s")
  print(f"  xB    : {xB:12.2f} m")
  print(f"  max_xB: {max_xB:12.2f} m")
  print(f"  vB    : {vB:12.2f} m/s{vB_rising}")
  if vB_turned_time is not None:
    print(f"  vBturn: {vB_turned_time:12.2f} s")


def main():
  simLoop()

main()


# EOF

Answer 8

Continuing on @hft's answer, let's try to solve $$ 0 = -gt - (\alpha v_0 + g)\frac{1}{\alpha}\left(e^{-\alpha t} - 1\right)\;. $$ In the Taylor series of that answer, we see terms of the form $v_0/g$ appearing. So to reduce the number of free parameters, let's introduce the "vacuum time" $t_v\equiv \frac{v_0}g$. The vacuum time is the amount of time to decelerate to zero in a vacuum, starting from $v_0$. In terms of this parameter we get

$$ 0 = -\frac{t}{t_v} - \left( 1 + \frac{1}{\alpha t_V}\right)\left(e^{-\alpha t} - 1\right)\;. $$ We cannot solve this equation, but Mathematica can. This gives $$t=\frac{\alpha t_v+W\left(-e^{\alpha (-t_v)-1} (\alpha t_v+1)\right)+1}{\alpha }$$ Here $W$ is the Lambert W function, or productlog in Mathematica, and is defined as the solution to $W(x) e^{W(x)}=x$. Plotting this expression gives

For a given $\alpha$, the time of flight $t$ looks a lot like a linear function in $t_v$. So double the initial velocity, double the time of flight$^\dagger$. As $\alpha$ increases, the slope of this almost linear line decreases, so the time of flight decreases for increasing $\alpha$. This seems to quickly asymptote as $\alpha$ gets larger. This is interesting, because at large $\alpha$ the initial speed doesn't seem to matter too much.

For high Reynolds numbers, which happen for small $\alpha$ and large $t_v$, this equation is expected to lose accuracy because then quadratic drag takes over.

$\dagger$ After some more playing, it seems that for constant $\alpha$ you get an expression that quickly asymptotes to $t=t_v+1/\alpha$.

Code to solve and create plot

sol = Solve[-  t/
     tv - (\[Alpha] + 1/tv) 1/\[Alpha] (Exp[-\[Alpha] t] - 1) == 0, t];
expr = t /. sol[[1]]
Plot3D[expr, {\[Alpha], 0, 5}, {tv, 0, 10}, 
 AxesLabel -> (Style[#, 14, Black] & /@ {\[Alpha], Subscript[t, v], 
     t})]

Answer 9

From the perspective of a schoolboy -- or as somebody with knowledge comparable to one --, and with due respect to the question which seems to me to be reasonable, the simplest and correct approach appears to be that of @Allure's diagram which might therefore be worth a closer look. First of all, surely the primary assumption of the question must be that whatever the 'air resistance', or 'drag force' actually is, it must apply symmetrically to the upward and downward motion of the ball moving in air, in which case the 'thin red line' in the diagram will be symmetric about the point at which it crosses the 'time' axis (t1): that is, symmetric about the line perpendicular to the tangent at that point.

As such, one's first inclination -- as a schoolboy to whom such an approach has presumably been demonstrated -- would be to regard the curve of the air ball's motion as that section of a hyperbola which, rather being asymptotic to the vertical 'velocity' axis, crosses it at the initial velocity point for both balls (t0), a hyperbola whose slope is equal to -- g at the point t1 at which v = 0 (viz, the tangent is parallel to the 'thick red line'), and which continues under that time axis to be asymptotic with the line representing its terminal velocity. Here it is necessary to consider that this effect that the air ball might reach such a velocity may distort the symmetry of the curve, but in the ordinary case within reasonable boundaries, this will not alter the outcome concerning which ball hits the ground first. The hyperbola seems apt moreover because its slope will be a function of -- 1/t^2, which means that if the 'air resistance' is the decelerating factor -- a, the distance travelled by the air ball will be proportional to the slope of the curve 1/t^2 (from Newton's laws, which all schoolboys know by heart).

The trick to the question is that, if the curve is indeed a hyperbola, then clearly the only way the distances of upward and downward motion for the air ball can be equal -- i.e. for the blue and green areas under the curve in the diagram to be equivalent -- is if the period between t1 and the final time t2, or t2 -- t1, is less than the period t1 -- t0, which means that since this latter period is already less than half the time taken for the vacuum ball to reach its apex (at v = 0), the total time t2 for the air ball to complete its flight is significantly less than that of the vacuum ball.

So, given the assumptions, principally the symmetry of the curve, the vacuum ball goes higher but the air ball lands first, always.

Also, while the inclination of 45 degrees of the 'thick red line' of slope -- g showing the motion of the vacuum ball is convenient, that angle may be arbitrarily chosen; the appropriate part of a hyperbolic curve symmetric about the point t1 may still be inserted.