Question

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?

Answer 1

The answer is 48.5 m

Step-by-step explanation:

First we need to know that in the y-axis there is acceleration, in specific the acceleration caused by the gravity. I consider this value as a 10 m/s2.

Then, we define the next equation for the y-axis:

`V_(f) ^(2) =  V_(i) ^(2) + 2gd`

Where:

Vi = 0 (because at the beginning there is no vertical velocity)

d = 0.027

Finally we have:

Vf = 0.7348 m/s

Also, we define another equation for the y-axis:

`V_(f) =  V_(i) + g*t`

we replace the variables and we have:

t = 0.07348 s

Finally, to calculate the horizontal distance, we need the next equation:

`D = v*t`

Replacing the variables we have as a result:

D = 48.5 m