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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
2 years, 6 months ago

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Take relative velocity along x and y axes and equate magnitude

Raghav Vaidyanathan - 2 years, 6 months ago

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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...

Abhineet Nayyar - 2 years, 6 months ago

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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.

Raghav Vaidyanathan - 2 years, 6 months ago

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

Abhineet Nayyar - 2 years, 6 months ago

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Yes i solved it . Any way thanx for helping .

Kyle Finch - 2 years, 6 months ago

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

Raghav Vaidyanathan - 2 years, 6 months ago

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Shouldn't be there plus sign in sin part

Kyle Finch - 2 years, 6 months ago

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Can u explain in brief

Kyle Finch - 2 years, 6 months ago

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