A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at 45o to the vertical. What is the speed of the rain? In what direction does rain fall as observed by a ground based observer?
A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at 45o to the vertical. What is the speed of the rain? In what direction does rain fall as observed by a ground based observer?

Suppose that Vrg is the velocity of the rain drop that appears to the female observer.

All of the vectors are drawn with reference to the frame from the ground up, save for one.

Let’s say the rain is falling at the following rate:

Case I

A girl’s velocity is equal to Vg in the context of the problem

Suppose that Vrg is the velocity of rain in relation to the girl.

The following is the equation that is used to determine a:

${v}_{r}–{v}_{g}=\left(a\stackrel{^}{i}+b\stackrel{^}{j}\right)–5\stackrel{^}{i}=\left(a–5\right)\stackrel{^}{i}+b\stackrel{^}{j}$

Therefore,

Case II

The girl’s velocity has increased as a result of this = vg

The following is the equation that is used to determine the value of b at 45o tan. It can be found here.

${v}_{r}–{v}_{g}=\left(a\stackrel{^}{i}+b\stackrel{^}{j}\right)–10\stackrel{^}{i}=\left(a–10\right)\stackrel{^}{i}+b\stackrel{^}{j}$

Therefore, the velocity of rain is:

${v}_{r}=5\stackrel{^}{i}–5\stackrel{^}{j}$

Speed of rain is:

$\left|{v}_{r}\right|=\sqrt{{5}^{2}+\left(–5{\right)}^{2}}=\sqrt{50}=5\sqrt{2}m/s$