For a man moving on track ABC, rain drops appear to fall vertically for the path AB, and along the same line of its motion (i.e., as observed by a stationary observer) for path BC. The actual speed of rain drop is (in m/s)

 [Step 1]:  $V_R=V_{x}i+V_{y}j$

[Step 2]: For path AB  $V_x=5m{s}^{-1}$
[Step 3]: For path BC velocity of rain  $\perp r$ to BC should be zero hence
[Step 4]:

 $\begin{array}{l}{V}_X\sin {30}^0=V_y\cos {30}^0\text{}5\times \frac{1}{2}=V_y\times \frac{\sqrt{3}}{2}\text{}\\ \Rightarrow V_y=\frac{5}{\sqrt{3}}m{s}^{-1}\end{array}$
[Step 5]:    $V_R=\sqrt{V_X^2+V_Y^2}=\sqrt{5^2+{\left(\frac{5}{\sqrt{3}}\right)}^2}=\sqrt{25+\frac{25}{3}}=\frac{10}{\sqrt{3}}m{s}^{-1}=5.77m{s}^{-1}$