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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TopNewestTake relative velocity along x and y axes and equate magnitude – Raghav Vaidyanathan · 2 years, 3 months ago

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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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– Raghav Vaidyanathan · 2 years, 3 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.Log in to reply

– Abhineet Nayyar · 2 years, 3 months ago

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

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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– Kyle Finch · 2 years, 3 months ago

Shouldn't be there plus sign in sin partLog in to reply

– Kyle Finch · 2 years, 3 months ago

Can u explain in briefLog in to reply

@Raghav Vaidyanathan @Tanishq Varshney @Ronak Agarwal – Kyle Finch · 2 years, 3 months ago

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