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,
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?