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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
1 year, 11 months ago

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Take relative velocity along x and y axes and equate magnitude Raghav Vaidyanathan · 1 year, 11 months ago

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@Raghav Vaidyanathan 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 · 1 year, 11 months ago

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@Abhineet Nayyar 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 · 1 year, 11 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 · 1 year, 11 months ago

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Yes i solved it . Any way thanx for helping . Kyle Finch · 1 year, 11 months ago

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\(v=u \sin {(\pi/4 -\theta)} \csc {(\pi/4 -\alpha)}\) Raghav Vaidyanathan · 1 year, 11 months ago

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@Raghav Vaidyanathan Shouldn't be there plus sign in sin part Kyle Finch · 1 year, 11 months ago

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@Raghav Vaidyanathan Can u explain in brief Kyle Finch · 1 year, 11 months ago

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