1. Would a spherical bullet fired horizontally from a musket stay aloft longer than a spherical bullet dropped from the same height?

    I know two similar questions have been asked (Will a bullet dropped and a bullet fired from a gun horizontally REALLY hit the ground at the same time when air drag is taken into account?, How can a horizontally fired bullet reach the ground the same time a dropped bullet does? ) but I want to narrow it down by making these assumptions:

    • No spin
    • Flat earth, no curvature
    • Spherical bullet

    It appears to me that the answer to my question is “yes” (and therefore the answer to the older related question is "no"), because aerodynamic force is more or less proportional to velocity squared. As the bullet starts to fall, at some given instant in time where the instantaneous velocity vector is known, if we take this velocity vector and thus compute the total aerodynamic force vector (which is purely a drag force vector; no lift is present) and then break this aerodynamic force vector into vertical and horizontal components, we get a larger vertical force component opposing downward acceleration than if we had used the same equation to calculate the aerodynamic drag force acting on an identical round ball falling straight down with the same instantaneous vertical velocity component but no forward motion. Is this correct?

    The same logic seems to shed light on why sideways winds exert much stronger sideways forces on moving cars than on parked cars, although here the situation is much more complicated because the rapidly moving car is acting somewhat like a vertical airfoil flying at an efficient angle-of-attack and creating sideways “lift”, while the parked car is more like the same airfoil in a completely “stalled” condition (due to the very high-- 90-degree-- angle-of-attack.) If the car were truly spherical, then this complication would be avoided and the situation would be just like the original question posed above. So,

  2. With a spherical car, would a given crosswind (blowing perpendicular to the road) exert a greater sideways force component on the car (i.e. a force component acting perpendicular to the road) if the car was driving forward than if the car were parked?

Let's give some numbers:

Ball falling at instantaneous speed 5 units, no forward motion, drag force is 25 units

Ball falling at instantaneous speed 5 units and moving forward at instantaneous speed 5 units, total instantaneous velocity vector magnitude 7.07 units, total drag force magnitude 50 units.

Vertical component of drag force is 50 units * sine 45 degrees = 35.4 units

Is this correct?

  • $\begingroup$ Here is a somewhat related question, but aerodynamic forces are ignored, so it's not ALL that related -- physics.stackexchange.com/q/479807 $\endgroup$ Jul 14, 2019 at 16:26
  • $\begingroup$ I'm tempted to edit to drop "lift" tag and add back one of original tags. Lift is only peripherally related here (i.e. in case of non-spherical car). In case of spherical bullet you only have a drag force, although you can convert that into vertical and horizontal components if you wish, each of which does include a component acting perpendicular to flight path. Really could be a whole other question whether is appropriate to introduce term "lift" here, in the case of spherical non-spinning bullet; I think not. $\endgroup$ Jul 14, 2019 at 18:52
  • $\begingroup$ Future edit: could change "vertical airfoil" to "sideways airfoil". $\endgroup$ Jul 14, 2019 at 19:03
  • $\begingroup$ @Qmechanic -- see second comment above. No big deal but maybe food for thought. Keeping in mind that the fundamental definition of lift is an aerodynamic force acting perpendicular to the flight path, i.e. perpendicular to the trajectory relative to the airmass. $\endgroup$ Jul 14, 2019 at 20:46
  • $\begingroup$ Hi quiet flyer. Welcome to Phys.SE. Please edit the tags if you think it would improve the question. $\endgroup$
    – Qmechanic
    Jul 14, 2019 at 21:07

2 Answers 2


In a vacuum they would fall at the same rate. Taking air drag into account, they probably would not. The early muskets that were not rifled were highly inaccurate at longer distances because they fired a round projectile with no spin on it. Any flaw, dent, or imperfection would cause the bullet to veer off in an unexpected direction because of uneven air flow. This is why rifled barrels were more accurate, if the bullet had a flaw and had more air drag on one side, since it was spinning it would travel in a spiral or wobble, in a much more predictable projectile arc. As for a spherical car. it should have very similar side drag whether moving or not. Moving forward it would have higher pressure at the front but less pressure at the rear so total sideways force should even out.

  • $\begingroup$ I understand the point about imperfections, but can you say something more specific about why the calculation at the end of my question is not correct? $\endgroup$ Jul 15, 2019 at 16:03
  • $\begingroup$ @quiet flyer: if I understand correctly in your calculation the ball would be moving downward 45 degrees from horizontal, and you are adding the 25 units of drag forward with 25 units of drag downward. But that would be incorrect, as some of the air resistance is added twice. in any direction of travel only the front half will create drag. $\endgroup$ Jul 15, 2019 at 19:20
  • $\begingroup$ can you explain more where you think my calculation went wrong? Yes the instantaneous direction of travel is 45 degrees from horizontal. I computed the total speed, then the total drag, then the vertical component of drag. Nothing got added twice. $\endgroup$ Jul 16, 2019 at 1:39
  • $\begingroup$ @quiet flyer: draw a circle to represent the ball, draw horizontal lines at top and bottom, ball is moving forward at 5 units, forward drag is 25 units. Now draw vertical lines at front and rear of the same circle, ball is dropping downwards at 5 units with a downward drag of 25 units. The ball is moving down and forward at 45 degrees. so your 25 plus 25 equals 50 units of drag. This represents 270 degrees of drag, not 180 $\endgroup$ Jul 16, 2019 at 1:58
  • $\begingroup$ I don't think it's correct to calculate the forward drag component (or the vertical drag component) before you've calculated the total drag. Take a look at the answer I provided and especially the first link embedded within it. Also I'm not following the significance of "270 degrees" in your last comment. $\endgroup$ Jul 16, 2019 at 2:02

I have now located a highly related answer that supports the calculation given at the end of the question, and supports the idea that the ball fired from the horizontally musket will stay aloft longer than the ball dropped from the same height, and also supports the idea that a crosswind will exert a stronger sideways force component on a spherical car that is driving forward than when the same car is parked. Here it is: Calculating wind force and drag force on a falling object

In truth the drag coefficient of a sphere cannot be considered strictly constant due to dependency of Reynolds number on speed, as pointed in another related answer Finding the drag force (Air resistance force) for accelerated ball? , but that doesn't appear to change the basic conclusion stated immediately above.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.