It depends on the initial height above the ground from which both projectiles are fired. I believe you can see that if I fire the vertical projectile from ground level (that is zero distance above the ground), it will travel some distance into the air. But if I fire the horizontal projectile from ground level, the projectile will travel zero distance before touching the ground (because it was already on the ground).
So, then let's say we lift the projectile a small distance $h$ above the ground, and then repeat the experiment. The vertical projectile will travel upward, being constantly decelerated by gravity until, finally, it reaches a velocity of zero. At this point it is at the apex of it's trajectory, some distance Δh above it's starting height. The horizontal projectile will also be acted upon by gravity, and as it moves horizontally, it will also be pulled down to Earth. It will continue to travel horizontally until gravity eventually pulls it to ground level, at which point it would have traveled some horizontal distance Δx.
Now, consider these two concepts:
1) The vertical projectile will always travel the same Δh no matter what height it is fired from. Say, it starts 5ft above the ground and reaches an apex at 10ft. Then, if it starts at 10ft and the same force is imparted it will reach an apex at 15ft. Either way, the change in vertical distance is the same. It is important to note that this is only true given that the height of the projectile remains fairly close to the Earth's surface. The force of gravity does become weaker as $h$ increases, but unless we are talking about very large values of $h$, the difference is negligible for this problem.
2) The horizontal projectile will travel a different Δx depending on the initial height it is fired from. Say, again, that it starts 5ft above the ground. It will have a certain amount of flight time before it reaches the ground. Now, if it is fired from 10ft above the ground, the amount of time it has to travel horizontally before it hits the ground is greater. Greater flight time, means more horizontal distance covered: greater Δx.
Therefore, if the vertical flight distance is independent of initial height, but the horizontal flight distance is dependent on initial height, that should tell you that the two distances will only be the same for select circumstances (if they are ever the same at all). In this case there is a specific relation between the initial magnitude of velocity of the projectiles to the projectiles' initial height above the ground that will allow for the two distances to be the same. It is possible to find the exact equation for the relationship, but I won't go into it here.
In other words, yes, the two distances can be the same under select circumstances, but by all means, they are not guaranteed to be the same. If one is interested in the mathematics behind these concepts, I would suggest picking up an intro level physics textbook
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!
Best Answer
Since I'm impatient I'll suggest one way you could be surprised: if
then the barrel is not level when aimed at a target the same height at the firing point, but instead points slightly upward accounting for the observation handily. Indeed, it must be that way because if the bullet was truly fired horizontally then it can only hit targets lower than the barrel.