Question

The speed of the rain falling vertically is$25\sqrt{3}\mathrm{m}/\sec$. The speed at which the wind is blowing from the west to the east is $25\mathrm{m}/\mathrm{s}$. Find the velocity of the rain experienced by the person standing on the ground.

Answer 1

Step 1: Given data

Speed of rain is $25\sqrt{3}\mathrm{m}/\sec$

Speed of wind is $25\mathrm{m}/\mathrm{s}$

Step 2: Formula used

If you have two vectors $\mathrm{A}$ and $\mathrm{B}$ as shown, the size of the resulting $\mathrm{C}$ will be:

$\mathrm{C}=\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{AB}\mathrm{cos\theta }}$

Here $\mathrm{\theta }$ is the angle between vectors $\mathrm{A}$ and $\mathrm{B}$.

Step 3: Determine the velocity of rain

${\mathrm{V}}_{\mathrm{r}}=25\sqrt{3}\mathrm{m}/\mathrm{s}$

${\mathrm{V}}_{\mathrm{w}}=25\mathrm{m}/\mathrm{s}$

The angle between these vectors is equal to $90°$. The resultant two vectors are given as,

$\mathrm{V}=\sqrt{{\mathrm{V}}_{\mathrm{r}}^2+{\mathrm{V}}_{\mathrm{w}}^2+2{\mathrm{V}}_{\mathrm{r}}{\mathrm{V}}_{\mathrm{w}}\cos 90°}$

$=\sqrt{{\mathrm{V}}_{\mathrm{r}}^2+{\mathrm{V}}_{\mathrm{w}}^2+0}$

$=\sqrt{(25\sqrt{3}{)}^2+(25{)}^2}$

$=25\sqrt{3+1}=25\sqrt{4}$

$=25\times 2=50\mathrm{m}/\mathrm{s}$

Therefore, the velocity of the rain is $50\mathrm{m}/\mathrm{s}$.