Answer
$ v_{0}=202\ 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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The angle is given in relation to the -y axis, and we need it relative to +x, so we write it as
$\theta=53^{o}-90^{o}=-37^{o}$
(diving at $37^{o}$ below the horizontal).
We are given $y_{0}=730m$ and $t=5.00s$ to reach $y=0.$
Use Eq. 4-22 to solve for $v_{0}:$
$y-y_{0}=(v_{0}\displaystyle \sin\theta_{0})t-\frac{1}{2}gt^{2}$
$ 0-730m=v_{0}\displaystyle \sin(-37.0^{o})(5.00s)-\frac{1}{2}(9.80m/\mathrm{s}^{2})(5.00s)^{2}$
$v_{0}=\displaystyle \frac{(-730m+\frac{1}{2}(9.80m/\mathrm{s}^{2})(5.00s)^{2}}{\sin(-37.0^{o})(5.00s)}$
$ v_{0}=202\ m/s.$
