A horizontal rifle is fired at a bull's-eye. The muzzle speed of the bullet is 745 m/s. The barrel is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.020 m below the center. What is the horizontal distance between the end of the rifle and the bull's-eye?
To find the horizontal distance between the end of the rifle and the bull's-eye, we need to consider the motion of the bullet in two dimensions: horizontal and vertical.
We are given the muzzle speed of the bullet (v₀ = 745 m/s) and the vertical displacement of the bullet (Δy = -0.020 m). We need to find the horizontal distance (Δx).
Let's assume that the time taken by the bullet to reach the target is t.
In the vertical direction, we can use the equation of motion:
Δy = v₀y * t + (1/2) * g * t²
Since the bullet is fired horizontally, the initial vertical velocity (v₀y) is zero. Also, the equation simplifies to:
Δy = (1/2) * g * t²
Solving for time, we can rearrange the equation:
t² = (2 * Δy) / g
Now, we know that the horizontal distance traveled by the bullet is given by:
Δx = v₀x * t
Since the bullet is fired horizontally, the horizontal velocity (v₀x) of the bullet remains constant throughout its motion, and it is equal to the muzzle speed.
Δx = v₀ * t
Now, we substitute the expression for t from the vertical equation into the horizontal equation:
Δx = v₀ * √[(2 * Δy) / g]
Finally, we can calculate the value of Δx using the given values:
Δx = 745 m/s * √[(2 * -0.020 m) / 9.8 m/s²]
Simplifying this expression gives:
Δx ≈ 3.05 m
Therefore, the horizontal distance between the end of the rifle and the bull's-eye is approximately 3.05 meters.