In a world without air, I understand they definitely would. However, with drag taken into account, I think they wouldn't. Since the drag force varies proportional to the square of speed (ignoring the change in coefficient of drag with speed), wouldn't the total impulse due to drag on a bullet fired from a gun from a specific height horizontal to ground be higher than the total impulse on an identical bullet dropped from the same height with no horizontal velocity?
Also, when bullet spin is taken into account, the bullet fired from a gun should resist change in orientation; therefore, it should maintain a small but non-zero angle of attack, and also have a lift force, correct?
Best Answer
Just based on the quadratic drag of air, yes, the fired bullet would take longer to hit the ground.
Just consider the vertical force caused by the air friction:
$F_y = - F_{\rm drag} \sin \theta = - C (v_x^2 + v_y^2) \frac{v_y}{\sqrt{v_x^2 + v_y^2}} = - C v_y \sqrt{v_x^2 + v_y^2}$
Where $\theta$ is the angle above the horizon for the bullet's velocity, and $C$ is some kind of drag coefficient. Note that when the bullet is moving down $\theta$ is negative, as is $v_y$, so the overall vertical force is positive and keeps the bullet off the ground for slightly longer.
In the dropped case, $v_x = 0$, so we get $F_y = -C v_y^2$.
In the fired case, we can neglect $v_y$ in the radical (assuming it's much smaller than $v_x$) and we get $F_y \approx -C v_y |v_x|$.
In other words, the upward force on the fired bullet is stronger, by a factor of $v_x / v_y$.
So freshman-level physics is wrong, at least according to sophomore-level physics.
Bonus Case:
If you're assuming a flat surface on earth, it's worth considering that many "flat" things (like the ocean) actually curve down and drop off below the horizon. In case you want to account for this curvature, it may be worth going to the bullet's reference frame with $\hat{y}$ always defined to point away from the center of the earth. Note that this puts you in a rotating reference frame, and then look at the centrifugal "force":
$F_y = m a = m R \omega^2 = m R \left(\frac{v_x}{R}\right)^2 = m \frac{v_x^2}{R} $
Where $R$ is the radius of the earth and $m$ is the mass of the bullet. So again, an upward force, this time proportional to $v_x$ squared. Of course this is the same as pointing out that the earth curves away from a straight line, but it's another fun application of not-quite-freshman physics.
Now you can add in much more complicated aerodynamics, but there the question sort of looses its undergrad physics charm there and becomes an aerospace engineering question!