\textbf{162-} A bullet is fired from a height of $95$ meters above the ground with initial speed $V_c$ at an angle of $37°$ above the horizontal, and after $5$ seconds it reaches a height of $40$ meters above the ground. The velocity vector of the bullet $3$ seconds after launch makes what angle with the launch velocity vector?
$$\left(g = 10\,\frac{\text{m}}{\text{s}^2},\quad \sin 37° = 0.6\right)$$
\begin{itemize}
\item[(1)] 37
\item[(2)] 45
\item[(3)] 53
\item[(4)] 90
\end{itemize}
\begin{center}
\textit{Calculation Space}
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