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The particle has initial velocity $\displaystyle \binom{v \cos \frac{\pi}{3}}{v \sin \frac{\pi}{3}}$ and acceleration $\displaystyle \binom{0}{-g}$. It will have zero vertical speed (ie be at the top of its trajectory) when $t = \frac{\sqrt{3}v}{2g}$.

Since $0 = v^2-u^2 + 2as$ the height achieved will be $\frac{3v^2}{8g}$

At this point it will need to travel the same distance again, but this time the initial speed is $\frac{3v}{2}$ so:

\begin{align*}
&& \frac{3v^2}{8g} &= \frac{3v}{2} t + \frac12 g t^2 \\
\Rightarrow && 0 &= 4g^2t^2+12vgt - 3v^2 \\
\Rightarrow && t &= \l \frac{-3+2\sqrt{3}}{2} \r \frac{v}{g}
\end{align*}

Therefore the total time is:

\begin{align*}
\l \frac{\sqrt{3}}{2} - \frac32 + \sqrt{3} \r \frac{v}{g} &= \frac{3\sqrt{3}-3}{2}\frac{v}{g}
\end{align*}

\begin{align*}
COM(\uparrow): && 0 &= 2m v_y - m \frac{3}{2}v \\
\Rightarrow &&v_y &= \frac34 v \\
COM(\rightarrow): && 3mV &= 2m (3v) +m \frac{v}{2} \\
\Rightarrow && V &= \frac{13}6\\
\end{align*}

Therefore superman is now travelling at a vector of $\displaystyle \binom{\frac{13}6}{\frac34}v$ ie an angle of $\tan^{-1} \frac 9{26}$ to the horizontal, approximately $19^\circ$

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