A horizontal rifle is fired at a bull's-eye. The muzzle speed of the bullet is 640 m/s. The gun is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.028 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 use the information given about the muzzle speed of the bullet and the vertical displacement.
First, let's consider the vertical motion of the bullet. We know that the bullet strikes the target 0.028 m below the center, which means it has a downward displacement of 0.028 m.
Using the known acceleration due to gravity (approximately 9.8 m/s^2) and the displacement, we can determine the time it takes for the bullet to reach the target using the following kinematic equation:
Δy = V₀y × t + (1/2) × a × t^2
Where:
Δy = vertical displacement = -0.028 m (negative because it's downward)
V₀y = initial vertical velocity = 0 (since the bullet is fired horizontally)
a = acceleration due to gravity = -9.8 m/s^2 (negative because it's downward)
t = time
Plugging in the values, we get:
-0.028 = 0 × t + (1/2) × (-9.8) × t^2
Simplifying the equation, we have:
-0.028 = -4.9t^2
Rearranging the equation, we get:
4.9t^2 = 0.028
Dividing both sides by 4.9, we obtain:
t^2 = 0.028 / 4.9
Taking the square root of both sides, we find:
t ≈ √(0.028 / 4.9)
t ≈ 0.065 s (rounded to three decimal places)
Now, let's calculate the horizontal distance traveled by the bullet in this time.
The horizontal distance traveled by the bullet can be given by the equation:
Δx = V₀x × t
Where:
Δx = horizontal distance traveled
V₀x = initial horizontal velocity (the muzzle speed of the bullet) = 640 m/s
t = time = 0.065 s (rounded value we calculated earlier)
Plugging in the values, we have:
Δx = 640 × 0.065
Δx ≈ 41.6 m (rounded to two decimal places)
Therefore, the horizontal distance between the end of the rifle and the bull's-eye is approximately 41.6 meters.