Answer
$12.0m.$
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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Given $v_{x}=v_{0}\cos 40.0^{\mathrm{o}}$ remains unchanged during flight, we use $\Delta x=22.0\mathrm{m}$ to find the time of flight:
$t=\displaystyle \frac{\Delta x}{v_{x}}=\frac{22.0\mathrm{m}}{(25.0\mathrm{m}/\mathrm{s})\cos 40.0^{\mathrm{o}}}=1.15\mathrm{s}.$
(a)
Equation 4-22 will give us the height above the release point at this time:
$\displaystyle \Delta y=(v_{0}\sin\sin\theta_{0})t-\frac{1}{2}gt^{2}$
$=(25.0m/s)\sin\displaystyle \sin 40.0^{o}(1.15s)-\frac{1}{2}(9.80m/\mathrm{s}^{2})(1.15s)^{2}$
$=12.0m.$
