# Question #8328e

Feb 3, 2018

While running up the hill,his velocity vector is $\left(2 i + 3 j\right) \frac{m}{s}$

And velocity of rain w.r.t the man is $- 4 j \frac{m}{s}$

So,velocity of rain w.r.t ground is $- 4 j + \left(2 i + 3 j\right) \frac{m}{s} = 2 i - j \frac{m}{s}$ (as velocity of rain w.r.t man = velocity of rain - velocity of man,so velocity of rain = velocity of rain w.r.t man + velocity of man)

While running down the hill,his velocity vector becomes $- \left(2 i + 3 j\right) \frac{m}{s}$

So,velocity of rain w.r.t the man while going down is = velocity of rain - velocity of man $= 2 i - j - \left(- 2 i - 3 j\right) = 4 i + 2 j \frac{m}{s}$

Hence the speed is $\sqrt{{4}^{2} + {2}^{2}} \frac{m}{s}$ i.e $\sqrt{20} \frac{m}{s}$