# Bullet meets Gravity

1. Oct 27, 2008

### nasjo30

1. The problem statement, all variables and given/known data
A horizontal rifle is fired at a bull's-eye. The muzzle speed of the bullet is 785 m/s. The barrel is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.029 m below the center. What is the horizontal distance between the end of the rifle and the bull's-eye?

2. Relevant equations
Kinematic Equations

3. The attempt at a solution
I tried creating a right triangle with the data but it didn't work and now im frustrated.

2. Oct 27, 2008

### fiziksfun

use the equation delta d = v(i)t + .5at^2

delta h (vertical) = .5at^2 where a is -9.81 m/s^2 [since v(i) in the vertical direction is 0)

-.029 = (.5)(-9.81)t^2

t=?

so you can solve for time and plug it into the equation again to get horizontal distance, right?

delta d (horizontal) = v(i)t [note, there is no acceleration in the horizontal direction)

where v(i) 785 m/s

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