Is the Book's Answer to the Relative Velocity Problem Incorrect?

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SUMMARY

The discussion centers on the relative velocity problem involving a man walking in the rain, where the book's answer contradicts the user's calculation. The user asserts that the relative velocity of rain to the man is represented by the equation R-M = -rJ - mI, leading to the conclusion that tan(theta) = 2/sqrt(5). However, the book provides the reciprocal answer of sqrt(5)/2. This discrepancy raises questions about the accuracy of the book's solution in multiple similar problems.

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manjuvenamma
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I raised this question in the courseware section and I got only one response. To improve the response, I am repeating the question here. Please see it.

I got one answer and the book at the receprocal of my answer. This is not just for this problem for several problems of this nature the answer given in the book is similarly reciprocal of mine. I want to just reconfirm who is consistently making this mistake.


1. Homework Statement
Rain is falling down vertically. To a man walking on the road, velocity of rain appears to be 1.5 times his velocity. Then to protect himself from rain, he holds his umbrella at an angle (theta) to the vertical such that tan (theta) =


2. Homework Equations

The relative velocity of rain to man is R-M where R is rain velocity and M is man's velocity.


3. The Attempt at a Solution

Assume the rain velocity vector is R. We can think it is -rJ. J is a unit vector along y axis. r is the magnitude of rain velocity. The negative sign comes because of the direction of rain - down.

Similarly M = mI where I is unit vector along x axis. The relative velocity of rain to man is R-M = -rJ-mI.

The magnitude of R-M is 1.5m (given).

1.5m = sqrt(r^2+m^2) implies m/r = 2/sqrt(5) = tan(theta).

But the answer stated in the book is the reciprocal i.e. sqrt(5)/2. Who is right?
 
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