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?
can someone solve?
how long does it take for the bullet to fall .020m?
h= 1/2 g t^2
now, having time t,
distance=horizontalvelocity*time
solve for distance.
What does the h you use stand for?
To solve this problem, we can use the equations of motion for projectile motion. First, let's break down the vertical and horizontal components of the bullet's motion.
1. Vertical Motion:
The bullet is fired vertically downwards with an initial vertical velocity of 0 m/s. The acceleration due to gravity (g) is -9.8 m/s² (taking downwards as negative). The displacement in the vertical direction is given as -0.020 m (negative because it is below the center).
Using the equation of motion:
Δy = v₀y * t + (1/2) * g * t²
Since the initial vertical velocity (v₀y) is 0, the equation simplifies to:
Δy = (1/2) * g * t²
Substituting the known values:
-0.020 m = (1/2) * (-9.8 m/s²) * t²
2. Horizontal Motion:
The bullet's horizontal velocity remains constant throughout the motion and is given as the muzzle speed of 745 m/s. The horizontal displacement (Δx) is what we need to find.
We can calculate the time taken (t) using the vertical motion equation:
Δy = (1/2) * g * t²
Solving for t:
t² = -2 * Δy / g
t = √(-2 * Δy / g)
Now, we can find the horizontal distance:
Δx = v₀x * t
Substituting the known values:
Δx = 745 m/s * (√(-2 * (-0.020 m) / -9.8 m/s²))
Simplifying the equation:
Δx = 745 m/s * √(0.040 m / -9.8 m/s²)
Finally, calculate the horizontal distance:
Δx ≈ 0.173 m
Therefore, the horizontal distance between the end of the rifle and the bull's-eye is approximately 0.173 meters.