Question

A hunter aims horizontally at a target on the same level as his rifle. If the bullet leaves his rifle at a velocity of 154.6 m/s and hits 0.64 meters below the target, how far away is the target? (hint: you are looking for dx)

Round your answer to the nearest 0.01. Do not add units.

Answer 1

The bullet's position vector has components

`x=v_0t`

`y=-\frac g2t^2`

where we're given the initial velocity
`v_0` and the initial height of the bullet
`y_0`, and
`g` is the acceleration due to gravity. Note that we take the bullet's starting position to be the origin.

We're also told that after some time
`t`, the time it takes for the bullet to hit the target, the bullet's vertical position is 0.64 m below where it was aimed, which translates to
`y=-0.64\,\mathrm m` at this time
`t`. Then we can solve for this time:

`-0.64\,\mathrm m=-\frac{9.80\,(\mathrm m)/(\mathrm s^2)}2t^2\implies t=0.36\,\mathrm s`

Finally, the target's distance from the origin is
`x` at this time
`t`:

`x=\left(154.6\,(\mathrm m)/(\mathrm s)\right)(0.36\,\mathrm s)=55.87\,\mathrm m`