# I 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?

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#### A.T.

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• ChessEnthusiast

#### jbriggs444

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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.

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• ChessEnthusiast

#### 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.

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