Finding the wind-resistance force

[ChessEnthusiast]

Imagine such situation:

There is a sniper who fires his rifle, the bullet travels at the initial speed of, say, 1400 m/s. His target is standing 2000 meters away and the wind is blowing at 30 m/s opposing the bullet's motion. Let's assume that the bullet is fired in a straight lane and it may fall the maximum of 0.5 m to still hit the target.
After how many seconds (if at all) will the target be hit?
Let's also assume that the cross-sectional area and the drag coefficient of the bullet is known.

Is there a way to calculate the force the wind will be opposing the motion of the bullet with?

 

[A.T.]

Is there a way to calculate the force the wind will be opposing the motion of the bullet with?

https://en.wikipedia.org/wiki/Drag_(physics)

 

[jbriggs444]

Imagine such situation:

There is a sniper who fires his rifle, the bullet travels at the initial speed of, say, 1400 m/s. His target is standing 2000 meters away and the wind is blowing at 30 m/s opposing the bullet's motion. Let's assume that the bullet is fired in a straight lane and it may fall the maximum of 0.5 m to still hit the target.
After how many seconds (if at all) will the target be hit?
Let's also assume that the cross-sectional area and the drag coefficient of the bullet is known.

Is there a way to calculate the force the wind will be opposing the motion of the bullet with?

Even ignoring air resistance, we have a problem. 2000 meters at 1400 meters per second will take longer than one second. In one second, a bullet will drop 4.9 meters ($\frac{1}{2}gt^2$ where g = 9.8 m/sec²) due to gravity. But you have imposed the requirement of a maximum 0.5 meter drop.

Edit: Drop from a bore-sighted, straight-at-the-target launch angle may not be what you are after. One can correct for that by aiming high. Instead, you may be constrained by the arch -- how much higher the mid-point of the trajectory is than the midpoint of the straight-line path. Ideally, that difference is lower by a factor of four. But that is still well over 0.5 meters.

 

[Zaya Bell]

If by straight line,l you mean horizontally, then range R=ut where u is initial velocity and t is time. Since the wind is directly against u, then you use vector addition. V= u+(-s) = 1400–30= 1370, where s is velocity of the wind. Now, in a projectile motion V(horizontal velocity) is constant neglecting drag. However, drag force is involved, therefore F=1/2C¶AV^2 where C is drag coefficient, ¶ is air density, and A is cross-sectional area. Remember that V was to move in a constant motion but since F is involved, it brings it to rest gradually.
Then F=(mV–m0)/t° = mV/t°. The time t can be computed.
Now from R=ut, 2000=1370t, t=2000/1370(time without drag force)
Finally, if t° >> t, or t°<t, then it can't hit the target. The reason is because

1) drag force reduces as the velocity reduces, making its calculation more complex. Nevertheless, this calculation should be a good estimation.​

2) The velocity turning to Zero is impractical since gravity will pull the bullet down to make impact with the ground.
For the force of the wind on the bullet,
F=m(2000–1370)/t`, if only we knew t`. This is different from wind resistance. The wind resistance(Air Friction) is just the drag force.

But it seem that your parameters are inconsistent and erroneous, however, assumed they were correct.