Ahorizontal rifle is fired at a bull's-eye. the muzzle speed of the bullet is 700 m/s. the gun is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.023 m below the center. what is the horizontal distance between the end of the rifle and the bull's-eye?

The horizontal distance between the end of the rifle and the bull´s-eye is 48.3 m

Explanation:

Please, see the attached figure for a description of the problem.

Notice that the vector r, that describes the position of the bullet when it reaches the target, is composed of rx and ry (see figure). Then:

r = (rx, ry)

We know that ry is -0.023 m (the reference system is located at the end of the rifle), then:

r =(rx, -0.023 m)

The equation for the vertical position of the bullet at time t is as follows:

y = y0 +v0y · t + 1/2 · g · t²

Since the reference system is located at the end of the rifle, y0 = 0. The vertical component of the speed at t = 0 is 0, then, v0y = 0.

Then, we can calculate the time at which the vertical position of the bullet is -0.023 m:

-0.023 m = 0m + 0m/s ·t -1/2 · 9.8 m/s² · t²

-0.023 m/ -4.9 m/s² = t²

t = 0.069 s

The equation for the horizontal position is given by this expression:

x = x0 + v0x · t

Since the origin of the reference system is located where the bullet starts its trajectory, x0 = 0. To know the horizontal distance between the bull´s-eye and the end of the rifle, we have to find the horizontal position of the bullet at the time at which its vertical position is -0.023 m (0.069 s)

Then:

x = v0x · t

x = 700 m/s · 0.069 s = 48.3m

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