A horizontal rifle is fired at a bull's-eye. The muzzle speed of the bullet is 660 m/s. The gun is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.027 m below the center. What is the horizontal distance between the end of the rifle and the bull's-eye?
See Related Questions: Fr1, 10-17-14, 9:38 PM.
To find the horizontal distance between the end of the rifle and the bull's-eye, we need to consider the horizontal motion of the bullet.
Let's assume that the bullet takes time 't' to reach the bull's-eye when fired horizontally. During this time, the bullet travels a horizontal distance 'd'.
Using the equation of motion for horizontal motion, we have:
d = v * t
where 'v' is the horizontal velocity of the bullet and 't' is the time of flight.
To find the time of flight, we can use another equation:
y = (1/2) * g * t^2
where 'y' is the vertical displacement of the bullet (0.027 m below the center), and 'g' is the acceleration due to gravity (approximately 9.8 m/s^2).
Rearranging the equation to solve for 't', we get:
t = sqrt((2 * y) / g)
Substituting the given values, we have:
t = sqrt((2 * 0.027) / 9.8)
t ≈ 0.0827 s
Now, let's substitute the value of 't' into the equation for horizontal distance 'd':
d = v * t
Substituting the given value of horizontal velocity 'v' (660 m/s) and time 't', we get:
d = 660 * 0.0827
d ≈ 54.342 m
Therefore, the horizontal distance between the end of the rifle and the bull's-eye is approximately 54.342 meters.