Answer
$h=51.8$ m
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
$\left[\begin{array}{ll}
x-x_{0}=(v_{0}\cos\theta_{0})t & (4- 21)\\
y-y_{0}=(v_{0}\sin\theta_{0})t-\frac{1}{2}gt^{2} & (4- 22)\\
v_{y}=v_{0}\sin\theta_{0}-gt & (4- 23)\\
v_{y}^{2}=(v_{0}\sin\theta_{0})^{2}-2g(y-y_{0}) & (4- 24)
\end{array}\right]$
$(a)$
The height $y=h$ was achieved after $t=5.50s,$
with
$y_{0}=0, v_{0}=42.0m/s, \theta_{0}=60.0^{o}.$
We find h using $4-22$
$h=y_{0}+v_{0}\displaystyle \sin\theta_{0}t-\frac{1}{2}gt^{2}$
$h=0+(42.0\displaystyle \mathrm{m}/\mathrm{s})(\sin 60^{o})(5.50s)-\frac{1}{2}(9.80m/s^{2})(5.50s)^{2}$
$h=51.8$ m
