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# Help again

A man is running up an incline planr maaking an angle $$\theta$$ with horizontal with a speed u .Rain drops falling at an angle $$\alpha$$ with the vertical appear to the man as if they are falling at an angle of $$45^o$$ with the horizontal. The speed of rain drops is

Note by Kyle Finch
1 year, 9 months ago

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Take relative velocity along x and y axes and equate magnitude · 1 year, 9 months ago

While talking about the rain, you took its direction in the south west direction. What prompted you to do that?? Why didn't you take it in the south-east direction?? In that case the x-component of relative velocity will be

ucos(theta)-vsin(alpha) and not ucos(theta)+vsin(alpha) as in this case... · 1 year, 9 months ago

Both are equivalent. Just interchange $$\alpha$$ in my equation with $$-\alpha$$ to get to the result that you have mentioned. This disparity arises since I am measuring the angle in the clockwise direction, while you are measuring it in the anticlockwise direction. · 1 year, 9 months ago

Yea, but the final expressions for the velocity are completely different...Then how are the results equivalent?? · 1 year, 9 months ago

Yes i solved it . Any way thanx for helping . · 1 year, 9 months ago

$$v=u \sin {(\pi/4 -\theta)} \csc {(\pi/4 -\alpha)}$$ · 1 year, 9 months ago

Shouldn't be there plus sign in sin part · 1 year, 9 months ago

Can u explain in brief · 1 year, 9 months ago