A horizontal rifle is fired at a bull's-eye. The muzzle speed of the bullet is 720 m/s. The barrel is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.031 m below the center. What is the horizontal distance between the end of the rifle and the bull's-eye?
h = 0.5g*t^2 = 0.031 m
4.9t^2 = 0.031
t^2 = 0.00633
Tf = 0.0795 s. = Fall time.
Dx = Xo*Tf = 720 m/s * 0.0795s = 57.3 m.
To find the horizontal distance between the end of the rifle and the bull's-eye, we first need to determine the time it takes for the bullet to travel to the target. We can do this by using the vertical motion of the bullet.
We know that the bullet strikes the target 0.031 m below the center, which means it traveled vertically downwards.
Using the equation of motion for vertical motion:
h = ut + (1/2)gt^2
Where:
h = vertical displacement (0.031 m down)
u = initial vertical velocity (0 m/s since the bullet is not initially moving vertically)
t = time
g = acceleration due to gravity (approximately 9.8 m/s^2)
Plugging in the values:
0.031 = 0t + (1/2)(9.8)t^2
Simplifying the equation:
4.9t^2 = 0.031
Dividing both sides by 4.9:
t^2 = 0.00632653
Taking the square root of both sides:
t ≈ 0.0796 seconds
Now that we have the time it takes for the bullet to travel, we can find the horizontal distance using the horizontal motion of the bullet. Since the barrel is pointed directly at the center of the bull's-eye, there is no initial horizontal velocity component.
The horizontal distance traveled by the bullet can be calculated using the equation:
d = vt
Where:
d = horizontal distance
v = horizontal velocity (muzzle velocity of the bullet)
t = time (0.0796 seconds)
Plugging in the values:
d = 720 m/s * 0.0796 s
Calculating the result:
d ≈ 57.47 meters
Therefore, the horizontal distance between the end of the rifle and the bull's-eye is approximately 57.47 meters.