# The biathlete fires at a vertical target from a distance of 200m. The barrel of his rifle is directed horizontally

**The biathlete fires at a vertical target from a distance of 200m. The barrel of his rifle is directed horizontally, the muzzle velocity of his bullet is 800m / s. The bullets hit the exact center of the target. How far from the center of the target will the bullet deviate, the initial velocity of which is 700 m / s due to the damp powder in the cartridge?**

S = 200 m.

g = 10 m / s2.

V1 = 800 m / s.

V2 = 700 m / s.

d -?

During the flight, the bullet is acted upon by the force of gravity m * g, directed vertically downward. The horizontal axis of the bullet out of the barrel must be above the center of the target at a distance d1.

Let us find the time of flight of the bullet t1 in the first case by the formula: t1 = S / V1.

d1 = g * t1 ^ 2/2 = g * S ^ 2/2 * V1 ^ 2.

Let us find the deviation from the bullet exit axis to the point of its hit for the second case d2.

The bullet flight time t2 is expressed by the formula: t2 = S / V ^ 2.

d ^ 2 = g * t2 ^ 2/2 = g * S ^ 2/2 * V2 ^ 2.

d = d ^ 2 – d1 = g * S ^ 2/2 * V2 ^ 2 – g * S ^ 2/2 * V1 ^ 2.

d = 10 m / s2 * (200 m) ^ 2/2 * (700 m / s) ^ 2 – 10 m / s2 * (200 m) ^ 2/2 * (800 m / s) ^ 2 = 0, 0955 m.

Answer: the bullet will deviate from the center of the target downward at a distance of d = 0.0955 m.