I don't deal with drag forces often, but I think the equation for drag is

$$F_D=Cv^2,$$

where $F_D$ is in the same direction as $v$, and $C$ contains all the various things - density of air, cross-section, drag coefficient, etc. Importantly, $C$ depends on the orientation of the object. What I am going to do is assume the bullet falls without rotating - so it stays parallel to the ground during its entire motion (in both cases - you drop it in the same direction you shoot it).

In the first case the equation of motion is found via Newton's second law:

$$\Sigma F_y=F_{D,y}-F_g=ma_y\rightarrow a_y=\frac{C_yv_y^2}{m}-g$$

In the second case, we need to consider both directions:

$$\Sigma F_y=F_{D,y}-F_g=ma_y\rightarrow a_y=\frac{C_yv_y^2}{m}-g$$ $$\Sigma F_x=-F_{D,x}=ma_x\rightarrow a_x=-\frac{C_xv_x^2}{m}$$

So to find the time of flight of either case one would have to integrate the $y$ equation, but in either case it is the same. Therefore, the time of flight for these two situations is the same. But of course, I'm assuming the bullet does not rotate during it's motion.

If it did rotate, then the value of $C$ would be constant - it would be $C_x$, because that's the direction of motion - and $F_D$ would be in the direction of motion of the bullet, and $v$ would be the speed. In this case I believe the other answer would be correct, and it they would reach the ground at different times.

I don't deal with drag forces often, but I think the equation for drag is

$$F_D=Cv^2,$$

where $F_D$ is in the same direction as $v$, and $C$ contains all the various things – density of air, cross-section, drag coefficient, etc. Importantly, $C$ depends on the orientation of the object. What I am going to do is assume the bullet falls without rotating – so it stays parallel to the ground during its entire motion (in both cases you drop it in the same direction you shoot it).

In the first case the equation of motion is found via Newton's second law:

$$\Sigma F_y=F_{D,y}-F_g=ma_y\rightarrow a_y=\frac{C_yv_y^2}{m}-g$$

In the second case, we need to consider both directions:

$$\Sigma F_y=F_{D,y}-F_g=ma_y\rightarrow a_y=\frac{C_yv_y^2}{m}-g$$ $$\Sigma F_x=-F_{D,x}=ma_x\rightarrow a_x=-\frac{C_xv_x^2}{m}$$

So to find the time of flight of either case one would have to integrate the $y$ equation, but in either case it is the same. Therefore, the time of flight for these two situations is the same. But of course, I'm assuming the bullet does not rotate during its motion.

If it did rotate, then the value of $C$ would be constant – it would be $C_x$, because that's the direction of motion – and $F_D$ would be in the direction of motion of the bullet, and $v$ would be the speed. In this case I believe the other answer would be correct, and they would reach the ground at different times.

I don't deal with drag forces often, but I think the equation for drag is

$$F_D=Cv^2,$$

where $F_D$ is in the same direction as $v$. $C$ contains all the various things - density of air, cross-section, drag coefficient, etc. So, in the first case the equation of motion is found via Newton's second law:

$$\Sigma F_y=F_{D,y}-F_g=ma_y\rightarrow a_y=\frac{Cv_y^2}{m}-g$$

In the second case, we need to consider both directions:

$$\Sigma F_y=F_{D,y}-F_g=ma_y\rightarrow a_y=\frac{Cv_y^2}{m}-g$$ $$\Sigma F_x=-F_{D,x}=ma_x\rightarrow a_x=-\frac{Cv_x^2}{m}$$

So to find the time of flight of either case one would have to integrate the $y$ equation, but in either case it is the same. Therefore, the time of flight for these two situations is the same.

I don't actually know enough about drag to say that the "v" in the drag equation is certainly, definitely the component of velocity and not the magnitude of the velocity, but if it WAS the magnitude, I think we would have a contradiction. If we dropped the bullet and asked "what is the force of drag in the horizontal direction?" and said "Cv^2", then a dropped bullet would spontaneously start moving in the horizontal direction. This is the same reason why surfaces can have different drag coefficients $C_i$ for different directions - the velocity in that direction is what determines the drag.

I don't deal with drag forces often, but I think the equation for drag is

$$F_D=Cv^2,$$

where $F_D$ is in the same direction as $v$, and $C$ contains all the various things - density of air, cross-section, drag coefficient, etc. Importantly, $C$ depends on the orientation of the object. What I am going to do is assume the bullet falls without rotating - so it stays parallel to the ground during its entire motion (in both cases - you drop it in the same direction you shoot it).

In the first case the equation of motion is found via Newton's second law:

$$\Sigma F_y=F_{D,y}-F_g=ma_y\rightarrow a_y=\frac{C_yv_y^2}{m}-g$$

In the second case, we need to consider both directions:

$$\Sigma F_y=F_{D,y}-F_g=ma_y\rightarrow a_y=\frac{C_yv_y^2}{m}-g$$ $$\Sigma F_x=-F_{D,x}=ma_x\rightarrow a_x=-\frac{C_xv_x^2}{m}$$

So to find the time of flight of either case one would have to integrate the $y$ equation, but in either case it is the same. Therefore, the time of flight for these two situations is the same. But of course, I'm assuming the bullet does not rotate during it's motion.

If it did rotate, then the value of $C$ would be constant - it would be $C_x$, because that's the direction of motion - and $F_D$ would be in the direction of motion of the bullet, and $v$ would be the speed. In this case I believe the other answer would be correct, and it they would reach the ground at different times.