Answer:
The horizontal distance between the end of the rifle and the bull´s-eye is 48.3 m
Step-by-step 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