Question

A man falling vertivally by parachute in a steady downpour of rain observes that when his speed is V1 the rain appears to make angle α with the upward vertical.When his speed is V2(V2>V1) the rain appears to make angle ß with the upward vertical.

Show that the rain actually falls at an angle Θ with the vertical given by

(V2-V1)cotΘ=V2cotα-V1cotß.

Answer 1

The velocity of man is $\vec V_{1}=-V_1\vec j.$

Suppose the relative velocity of the wind is

$$\vec V_{WM_1}=a\sin(\alpha)\vec i+a\cos(\alpha)\vec j$$

The velocity of the wind is

$$\vec V_W=\vec V_{1}+\vec V_{WM_1}=a\sin(\alpha)\vec i+(-V_1+a\cos(\alpha))\vec j$$

The velocity of man is $\vec V_{2}=-V_2\vec j.$

Suppose the relative velocity of the wind is

$$\vec V_{WC2}=b\sin(\beta)\vec i+b\cos(\beta)\vec j$$

The velocity of the wind is

$$\vec V_W=\vec V_{2}+\vec V_{WC_2}=b\sin(\beta)\vec i+(-V_2+b\cos(\beta))\vec j$$

Then

$$\begin{matrix} a\sin(\alpha) =b\sin(\beta) \\ -V_1+a\cos(\alpha)=-V_2+b\cos(\beta) \end{matrix}$$

$$\begin{matrix} a =b\cdot\dfrac{\sin(\beta)}{\sin(\alpha)} \\ V_2-V_1=b\sin(\beta)(\cot(\beta)-\cot(\alpha)) \end{matrix}$$

$$\cot(\Theta)=\dfrac{-V_2+b\cos(\alpha)}{b\sin(\beta)}$$

$$b\sin(\beta)\cot(\Theta)=-V_2+b\sin(\beta)\cot(\beta)$$

$$(V_2-V_1)\cot(\Theta)=-V_2((\cot(\beta)-\cot(\alpha)))$$

$$+(V_2-V_1)\cot(\beta)$$

$$(V_2-V_1)\cot(\Theta)=V_2\cot(\alpha)-V_1\cot(\beta)$$