Ahorizontal rifle is fired at a bull's-eye. the muzzle speed of the bullet is 670 m/s. the gun is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.025 m below the center. what is the horizontal distance between the end of the rifle and the bull's-eye?

47.86 m

Explanation:

Consider the motion along the vertical direction and assume the motion in downward direction as negative

y = vertical displacement = - 0.025 m

a = acceleration = - 9.8 m/s²

t = time taken to strike the target below the center

= initial velocity along the vertical direction = 0 m/s

Using the kinematics equation

y = t + (0.5) a t²

Inserting the values

- 0.025 = (0) t + (0.5) (- 9.8) t²

t = 0.0714 sec

Consider the motion along the horizontal direction

x = horizontal displacement = horizontal distance between the end of the rifle and the bull's-eye

v₀ = velocity along the horizontal direction = 670 m/s

Since there is no acceleration along the horizontal direction, we have

x = v₀ t

inserting the values

x = (670) (0.0714)

x = 47.86 m

area of focus

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.