# 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
5 years, 6 months ago

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Sort by: Take relative velocity along x and y axes and equate magnitude

- 5 years, 6 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...

- 5 years, 6 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.

- 5 years, 6 months ago

Yea, but the final expressions for the velocity are completely different...Then how are the results equivalent??

- 5 years, 6 months ago

- 5 years, 6 months ago

$v=u \sin {(\pi/4 -\theta)} \csc {(\pi/4 -\alpha)}$

- 5 years, 6 months ago

Can u explain in brief

- 5 years, 6 months ago

- 5 years, 6 months ago

Yes i solved it . Any way thanx for helping .

- 5 years, 6 months ago