Answer
$| \vec{v_{avg}}|=5.8$ m/s.
Work Step by Step
If $\vec{v}_{0}$ is expressed as a magnitude (the speed $v_{0}$) and an angle $\theta_{0}$ (measured from the horizontal),
the particle's equations of motion along the horizontal $x$ axis and vertical $y$ axis are
$\left[\begin{array}{ll}
x-x_{0}=(v_{0}\cos\theta_{0})t & (4- 21)\\
y-y_{0}=(v_{0}\sin\theta_{0})t-\frac{1}{2}gt^{2} & (4- 22)\\
v_{y}=v_{0}\sin\theta_{0}-gt & (4- 23)\\
v_{y}^{2}=(v_{0}\sin\theta_{0})^{2}-2g(y-y_{0}) & (4- 24)
\end{array}\right]$
The initial velocity components are $\left\{\begin{array}{l}
v_{0x}=v_{0}\cos\theta_{0}\\
v_{0y}=v_{0}\sin\theta_{0}
\end{array}\right.$
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$\Delta y$=0 because both the initial and final heights are 0.\ We find the time of flight from 4-22.
$\displaystyle \Delta y=0=(v_{0}\sin\theta_{0})t-\frac{1}{2}gt^{2}$
$t=\displaystyle \frac{2v_{0}\sin\theta_{0}}{g}$
$t=\displaystyle \frac{2(19.5\mathrm{m}/\mathrm{s})\sin 45.0^{\mathrm{o}}}{(9.80\mathrm{m}/\mathrm{s}^{2})}=2.81s$
In this time, the horizontal distance covered is given by 4-21
$x-x_{0}=(v_{0}\cos\theta_{0})t=(19.5\mathrm{m}/\mathrm{s})\cos 45.0^{\mathrm{o}}(2.81s)=38.7$ m
This is in direction to to the player, (our +x direction)
The player needs to make a displacement of
$\Delta\vec{r}=(38.7m-55m)\hat{i}=(-16.3m)\hat{i}$
during $\Delta t=2.81s,$
So his average velocity is $\displaystyle \vec{v_{avg}}=\frac{\Delta\vec{r}}{\Delta t}=\frac{(-16.3m)\hat{i}}{2.81s}=(-5.80m/s)\hat{i}$
and his average speed is $| \vec{v_{avg}}|=5.80$ m/s.
