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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2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
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Top NewestTake relative velocity along x and y axes and equate magnitude
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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...
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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.
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Yes i solved it . Any way thanx for helping .
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\(v=u \sin {(\pi/4 -\theta)} \csc {(\pi/4 -\alpha)}\)
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Shouldn't be there plus sign in sin part
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Can u explain in brief
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@Raghav Vaidyanathan @Tanishq Varshney @Ronak Agarwal
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