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Objects take same time to fall a given height independent of their horizontal speed when air resistance is ignored, is this also true when air resistance is not ignored? In presence of air resistance, the drag co-efficient needs to be considered and also the fact that air resistance increases with increasing speed. So, if three identical objects are shot with three different initial speeds, won't they reach the ground at slightly different times? Can anyone please explain?

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  • $\begingroup$ Yes, Fig. 3.2 assumes no drag. $\endgroup$ – Qmechanic May 23 at 18:42
  • $\begingroup$ @Qmechanic Doesn't the time in the air depend only on vertical air drag and not horizontal air drag? That is, the statement in Fig 3.2 applies with or without air drag. I would think horizontal air drag only effects the horizontal distance traveled not the time in the air, but maybe I'm missing something. $\endgroup$ – Bob D May 23 at 19:08
  • $\begingroup$ The answer depends on the regime (that is, on the effective force law for drag). Under the right force law the equal time conditions still stands. The surprise is how close to that special force law our day-to-day experience hews. $\endgroup$ – dmckee May 23 at 19:18
  • $\begingroup$ @dmckee I'm looking at a NASA.gov site where they give the following equation for air drag- is this the "force law" you are referring to? They apply it to a falling object and no mention is made of needing to take into account possible horizontal drag.$$D=C_{d}ρ\frac{V^{2}A}{2}$$ Where $C_{d}$ is the drag coefficient, $ρ$ is the air density, $V$ is the downward velocity and $A$ is the reference (frontal) area $\endgroup$ – Bob D May 23 at 19:35
  • $\begingroup$ @Bob That is the Rayleigh drag which generally applies for Reynolds number above roughly 1000. And you should note that $C_d$ is in general a function of velocity (though for many shapes it is nearly constant over a significant ranges of Reynolds number). At lower and much higher Reynolds number other laws obtain (and things are strange around mach one). $\endgroup$ – dmckee May 23 at 19:52
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I am still not sure why there won't be any drag in this case, anyone can explain in layman's terms?

I will venture to give it try. There will be air drag both horizontally and vertically, but that air drag force is always in the direction opposing the direction of motion.

There are basically two forces acting on a projectile in air: Gravity and air drag (air resistance).

The air drag force is directed so as to oppose the direction of motion. Gravity is always directed downward.

The thing that affects air time (time to fall from the initial height to the ground) is the downward component of the acceleration of the projectile. The things that affect the downward acceleration is the downward force of gravity, $mg$ (the objects weight, which we will consider constant for the height involved) and the upward air drag force acting opposing its downward motion, which we can call $D$. So since $F=ma$ or $a=\frac{F}{m}$, the downward vertical acceleration of the object will be

$$a_{vertical}=-\frac{mg-D}{m}$$

Where $D$ is the upward air drag force. I need to emphasize that $D$ is not a constant. It involves many variables. Perhaps most importantly if is a function of velocity and can even vary as the square of the velocity. All other factors being being equal, the drag force increases with increasing velocity until the drag force equals the gravitational force and the velocity becomes constant (the so called terminal velocity).

There is no horizontal forces acting on the projectile other than horizontal air drag force which opposes its horizontal motion causing it to decelerate. Therefore the horizontal acceleration is

$$a_{horizontal}=\frac{-D}{m}$$

Bottom line. The horizontal and vertical accelerations, and therefore velocities, are basically independent of one another. The different initial horizontal velocities of the cannon balls should not affect the air time (falling time) of the cannon balls. They will, however, effect the distance travelled by the different balls.

In closing I suspect based on the comments going back and forth others may disagree with me. I have an open mind and am amenable to improving this answer based on sound technical input from others

Hope this helps.

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  • $\begingroup$ This is currently erroneous. You need trig functions on the drag, because the direction of the drag is against the current direction of motion. $a_\text{hor.} = -D \cos \theta_v / m$ where $\theta_v = \arctan (v_y/v_x)$. This is why, in general, drag couples the vertical and horizontal motion and why it is only for a special drag force law that the falling time for the two cases are identical. $\endgroup$ – dmckee May 24 at 18:55
  • $\begingroup$ @dmckee Then you need to tell me what the "special drag force" law is. Can't the the drag force be resolved into horizontal and vertical components? And only the vertical component should affect fall time? Please explain how the "special drag force law" changes the vertical acceleration . $\endgroup$ – Bob D May 24 at 20:01
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Of course drag means that any trajectory that we are considering in this problem will take longer than the corresponding trajectory in a drag-free environment.

But that fact alone does not tell you if the straight-drop and projected-horizontally trajectories will or will not require the same time to fall or not, only that each will take longer than the case where there was no drag.

To work out the time required we need only know how the vertical component $D_y$ of the drag $\vec{D}$ varies. Now, drag depends on the magnitude of velocity and always points opposite the direction of velocity so we can write \begin{align} \vec{D} &= - f(v)\hat{v} \\ &= - f(v) \frac{\vec{v}}{v} \end{align} for some (as yet unknown) function $f$, which leads to the form \begin{align} D_y &= - f(v) \frac{v_y}{v} \\ &= - \frac{f(v)}{v} v_y \\ &= g(v) v_y \;. \end{align} In order for the falling time to to independent of the initial horizontal velocity, we need $g(v)$ to be independent of $v_x$, which means it should be independent of $v$ (i.e. a constant of some kind). Which implies that the falling drag rule that preserves the equal-time condition for these options is of the form $D_\text{equal-time} = -k v$.


Explicit calculations

Rayleigh drag

The drag rule that applies for most day-to-day situations isthe Rayliegh drag $$ D_\text{Rayleigh} = -\frac{1}{2} \rho C_d A v^2 \hat{v} \;,$$ where $A$ is the cross-sectional area presented by the object, $\rho$ is the density of the fluid through which the projectile moves, and $C_D$ is a "coefficient of drag" expressing the dependence on shape and orientations (which is roughly but not exactly independent of speed for many interesting cases).

Consider a moment where the particle has velocity $\vec{v} = (v_x, v_y)$, the y-component of drag is \begin{align} D_y &= \frac{1}{2} \rho C_D A \, v \,v_y \\ &= \frac{1}{2} \rho C_D A \sqrt{v_x^2 + v_y^2} \,v_y \;. \end{align} which does depend on the horizontal velocity.

Stokes drag

At very low Reynolds number the dominate drag is the Stokes drag $$ D_\text{Stokes} = -6 \pi \mu R v \hat{v} \, $$ where $\mu$ is the dynamic viscosity of the fluid, and $R$ is a characteristic size of the system.

Here the vertical component of drag is \begin{align} D_y = 6 \pi \mu R v_y \, \end{align} and does not depend on the horizontal component of velocity.

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