Bullets fired, when can I hear it?

Simple: The time it took the game keeper to hear the bullet is 1/340s. Using the kinematics equation X=Vi t+(a t^2)/2 where there is no acceleration, as in this problem, we are left with X=Vi*t. plug in for t=1/340s and Vi=680m/s and the answer is x=2

Suppose the bullet is travelling along x-axis, following the function $x(t)=680t$ . We'll place the gamekeeper at the point (0,1) without loss of generality.

The time it takes for the sound of the bullet at some point $x$ to reach the gamekeeper is $\frac{1}{340}\sqrt{1+x^2}$ . We know the sound of the bullet at position $x$ starts at time $t$ , so $t+\frac{1}{340}\sqrt{1+(680t)^2}$ must be the equation that describes when the gamekeeper hears the bullet from point $x$ . The minimum value of this equation is the time when the gamekeeper first hears the sound of the bullet.

The minimum value is $\frac{\sqrt{3}}{680}$ , so the position of the bullet is $\sqrt{3}$ .

The distance formula yields us $\sqrt{1^2+\sqrt{3}^2}=\boxed{2}$ .