Answer
$\displaystyle \theta_{0}= 78.5^{\mathrm{o}}.$
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 $\left\{\begin{array}{l}
v_{0x}=v_{0}\cos\theta_{0}\\
v_{0y}=v_{0}\sin\theta_{0}
\end{array}\right.$
At maximum height, $v_{y}=0$, so the velocity at that point equals $\vec{v}=v_{0x}\hat{i}$,
because the horizontal component remains the same throughout the flight (there is no horizontal acceleration).
The speed at the maximum point is $v=v_{0}\cos\theta_{0}$, and this equals $\displaystyle \frac{1}{5}v_{0} $
(a fifth of the initial speed).
So we solve for $\theta_{0}$
$v_{0}\displaystyle \cos\theta_{0}=\frac{1}{5}v_{0}$
$\displaystyle \cos\theta_{0}=\frac{1}{5}$
$\displaystyle \theta_{0}=\cos^{-1}\frac{1}{5}=78.5^{\mathrm{o}}.$
