The horizontal distance between the end of the rifle and the eye of the bull is equal to 50.35 m.
The equations of motion can be described as the equations which set up the relationship between the velocity of the object, the acceleration, time, and distance of the moving object.
Consider that a vector r, which describes the position of the bullet when it reaches the target, is composed of [tex]r_x[/tex] and [tex]r_y[/tex]
[tex]r = (r_x, r_y)[/tex]
Given that [tex]r_y[/tex] is -0.029 m (the reference is located at the end of the rifle),
r =([tex]r_x[/tex], -0.029 m)
The vertical position of the bullet at time 't' can be written in form of an equation as follows:
[tex]y = (y_o +v_oy) t + \frac{1}{2} g t^2[/tex]
The vertical component of the speed at t = 0 is 0, then, acceleration, v₀y = 0 and y₀ = 0.
Then, we can find the time for the vertical position (y) of bullet -0.029 m:
-0.029 m =(0)t - (1/2) · (9.8 m/s²) · t²
t = 0.077 sec
The equation for the horizontal position can be written as:
x = x₀ + (v₀x) · t
Since the origin of the reference system is located at the bullet starts its trajectory, x₀ = 0.
x = (v₀x) · t
x = (655) × 0.077 = 50.35m
Learn more about equations of motion, here:
https://brainly.com/question/16982759
#SPJ1