A horizontal rifle is fired at a bull's-eye. The muzzle speed of the bullet is 692 m/s. The barrel is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.0239 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 this case, the bullet is being fired horizontally, which means there is no initial vertical velocity (Vy = 0). The only acceleration acting on the bullet is due to gravity, which causes it to fall below the straight-line path. Therefore, the bullet's vertical motion can be described by the equations of motion under constant acceleration.
Using the equation of motion for displacement in the vertical direction, we have:
Δy = Vy*t + (1/2)*ay*t^2
Since the initial vertical velocity is 0, the equation simplifies to:
Δy = (1/2)*ay*t^2
We are given the vertical displacement (Δy) of the bullet, which is 0.0239 m. Additionally, the acceleration due to gravity (ay) is approximately 9.8 m/s^2. We can rearrange the equation and solve for time (t):
0.0239 = (1/2)*9.8*t^2
Simplifying the equation further:
0.0239 = 4.9*t^2
Dividing both sides of the equation by 4.9:
t^2 = 0.0239 / 4.9
Taking the square root of both sides of the equation:
t ≈ sqrt(0.0239 / 4.9)
t ≈ 0.0867 s
Now, we have the time it takes for the bullet to reach the bull's-eye (t). To find the horizontal distance traveled by the bullet, we can use the equation of motion for displacement in the horizontal direction:
Δx = Vx*t
The muzzle speed of the bullet (Vx) is given as 692 m/s. We can substitute the values into the equation:
Δx = 692 * 0.0867
Δx ≈ 60.0134 m
Therefore, the horizontal distance between the end of the rifle and the bull's-eye is approximately 60.0134 meters.