Question #8328e

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Question #8328e

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Answer 1 · 2018-02-03T09:56:06

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

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

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

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

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

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Question #8328e

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