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
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Kyle Finch
2 years, 3 months ago
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Top NewestTake relative velocity along x and y axes and equate magnitude – Raghav Vaidyanathan · 2 years, 3 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, 3 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, 3 months ago
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Yea, but the final expressions for the velocity are completely different...Then how are the results equivalent?? – Abhineet Nayyar · 2 years, 3 months ago
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Yes i solved it . Any way thanx for helping . – Kyle Finch · 2 years, 3 months ago
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\(v=u \sin {(\pi/4 -\theta)} \csc {(\pi/4 -\alpha)}\) – Raghav Vaidyanathan · 2 years, 3 months ago
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Shouldn't be there plus sign in sin part – Kyle Finch · 2 years, 3 months ago
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Can u explain in brief – Kyle Finch · 2 years, 3 months ago
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@Raghav Vaidyanathan @Tanishq Varshney @Ronak Agarwal – Kyle Finch · 2 years, 3 months ago
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