Why two doppler shifts when reflecting from a moving object? 
A baseball coach uses a radar device to measure the speed of an
  approaching pitched baseball. This device sends out electromagnetic
  waves with frequency $f_0$, and then measures the shift in frequency
  $\Delta f$ of the waves reflected from the moving baseball. If the
  fractional frequency shift produced by a baseball is $\frac{\Delta
f}{f_0}=2.86\times10^{-7}$, what is the baseball's speed? (Hint: Are the
  waves Doppler-shifted a second time when reflected off the ball?)

I know that because the ball has a non-relativistic velocity, there are certain terms of the Doppler effect equation that can be simplified, such that $\frac{\Delta f_0}{f}=\frac{u}{c}$, but I do not understand the relation that exists with the waves reflected in the ball. 
The answer to the problem is $u=\frac{\Delta f_0}{2f}(c)=\frac{2.86\times 10^{-7}}{2}(3\times10^8\,{\rm m})=42.9\,{\rm m}\,{\rm s}^{-1}=154\,{\rm km}\,{\rm h}^{-1}$
Where does the "$2$" in $u=\frac{\Delta f_0}{2f}(c)$ of the answer come from?
 A: Another way of visualizing the need for a factor of two (which amounts to the same thing as the answer of @Now IGet...,):
As far as the Doppler shift is concerned, the ball is, in effect, a plane mirror.  It is forming a virtual image of the radar gun (as a source) behind it;  this virtual source has its velocity measured by the radar gun. The distance from the ball to the real source always equals the distance from the ball to the virtual source.
When the ball moves a distance $x$ towards the real detector, the virtual detector also moves a distance $x$ towards the ball, or a distance $2x$ towards the real gun.
So, after using the standard Doppler shift equation to find the velocity of the virtual source towards the detector, you need to divide the answer by $2$ to find the velocity of the ball...
A: There is a factor of two because there are two doppler shifts: one when the ball sees the wave from the coach, and one when the coach sees the waves reflected by the moving ball. The baseball sees a fractional frequency shift of $u/c$ owing to the balls own motion. Let's say for concreteness that the radar gun emitted a flash of light every second, and $u/c$ is $10\%$.  Then the ball sees flashes of light every $0.9$ seconds, since it is moving towards the coach. Therefore the ball is relecting light with a frequency shifted up by $u/c$, so every $0.9$ seconds. 
But now the coach is seeing this signal coming from a moving source, so there is another doppler shift owing to the fact that the baseball is a moving source. The fractional shift is again $u/c$, so even though the ball is flashing every $0.9$ seconds, the coach sees  a flash every $0.8$ seconds, so you get a total fractional shift of $2u/c$, so $\Delta f/f_0=2u/c$.
