Question

На какое расстояние в горизонтальном направления отклонится пуля, масса которой 6 г, под действием бокового ветра, действующего с силой 1,2 x 10 в минус второй степени Н? Начальная скорость пули 600 м / с, дистанция стрельбы 300м, угол между направлением стрельбы и ветра 90 градусов. Скорость пули в направлении стрельбы считать постоянной

Answer 1

F=m*a => a=F/m
t=l/v
y=a*t²/2 - смещения
решая систему получаем y=0.25 м.

Answer 2

Calculation of Bullet Deviation Due to Crosswind

To calculate the horizontal deviation of the bullet due to the crosswind, we can use the formula:

Horizontal Deviation = (Wind Force * (Time of Flight)^2) / (2 * Bullet Mass)

Where: - Wind Force = 1.2 x 10^-2 N - Bullet Mass = 6 g = 0.006 kg - Initial Bullet Velocity = 600 m/s - Distance of Shot = 300 m

First, we need to calculate the time of flight of the bullet. We can use the formula:

Time of Flight = Distance / (Initial Bullet Velocity * cos(θ))

Where: - θ = 90 degrees (angle between the direction of the shot and the wind)

Calculating the time of flight: Time of Flight = 300 / (600 * cos(90))

Using the trigonometric identity cos(90) = 0, the denominator becomes 0, which means the bullet will not experience any horizontal deviation due to the crosswind.

Therefore, the horizontal deviation of the bullet due to the crosswind is 0 meters.

Conclusion

Given the parameters provided, the horizontal deviation of the bullet due to the crosswind is 0 meters. This is because the angle between the direction of the shot and the wind is 90 degrees, resulting in a time of flight of 0 seconds and no horizontal deviation.