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

$x=v_0t$

$y=-\dfrac g2t^2$

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:

$-0.64\,\mathrm m=-\dfrac{9.80\,\frac{\mathrm m}{\mathrm s^2}}2t^2\implies t=0.36\,\mathrm s$

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

$x=\left(154.6\,\dfrac{\mathrm m}{\mathrm s}\right)(0.36\,\mathrm s)=55.87\,\mathrm m$