163. In the figure, a weight is thrown upward from the bottom of an inclined surface with initial velocity $V_0$, tangent to the surface. The weight goes up and returns to the starting point. If the friction force is $\frac{2}{10}$ of the weight, the time to go up equals the time to come down by how much? ($g = 10\,\frac{m}{s^2}$) [Figure: inclined surface at $30^\circ$ with initial velocity $V_0$ along the surface] (1) $\sqrt{\dfrac{4}{3}}$ (2) $\sqrt{\dfrac{3}{7}}$ (3) $\dfrac{3}{5}$ (4) $\dfrac{5}{3}$
\textbf{163.} In the figure, a weight is thrown upward from the bottom of an inclined surface with initial velocity $V_0$, tangent to the surface. The weight goes up and returns to the starting point. If the friction force is $\frac{2}{10}$ of the weight, the time to go up equals the time to come down by how much? ($g = 10\,\frac{m}{s^2}$)
\textit{[Figure: inclined surface at $30^\circ$ with initial velocity $V_0$ along the surface]}
\hspace{1cm} (1) $\sqrt{\dfrac{4}{3}}$ \hspace{2cm} (2) $\sqrt{\dfrac{3}{7}}$ \hspace{2cm} (3) $\dfrac{3}{5}$ \hspace{2cm} (4) $\dfrac{5}{3}$