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Sir my relative motion is somehow weak ,so plz explain me the concept of way of solving these type of problem,Thanks Sir
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Asked by vishakhachandan026 | 21 May, 2019, 07:19: AM
answered-by-expert Expert Answer
Fig.1 describes the method of getting relative velocity.
Let B moves with velocity vB and A moves with velocity vA as shown in fig.1 .
 
Let θ be the angle between the velocity directions of A and B.
Relative velocity of B with respect to A  is obtained from the diagonal of parallelogram
formed by vectors -vA and vB as shown in figure 1.
Angle Φ made by relative velocity vector with -vA is given by,   begin mathsize 12px style tan space ϕ space equals space fraction numerator v subscript B sin theta over denominator v subscript A minus space v subscript B cos theta end fraction end style .................................(1)
 
In the given problem we need to get relative velocity of wind with respect to launcher because the flag in the
launcher will point in the direction of relative velocity of wind.
 
Fig.2 shows the vector diagram , showing the velocity V of launcher , velocity u of wind and
relative velocity of wind (u-V) with respect to launch.
 
Angle between velocity vector V of launcher and velocity u of wind is (90-α). 
Angle between relative velocity vector (u-V) and -V is (180-β).
 
Hence as per eqn.(1), we write,    begin mathsize 12px style tan left parenthesis 180 minus beta right parenthesis space equals space fraction numerator u space sin open parentheses 90 minus alpha close parentheses over denominator v minus u space cos open parentheses 90 minus alpha close parentheses end fraction space end style
begin mathsize 12px style negative space tan beta space equals space fraction numerator u space cos alpha over denominator V minus u space sin alpha end fraction V space equals space u space sin alpha space minus space fraction numerator u space cos alpha over denominator tan space beta end fraction space space equals space fraction numerator u left parenthesis sin alpha space sin beta space minus space cos alpha space cos beta right parenthesis over denominator sin beta end fraction space equals space minus space fraction numerator u space cos open parentheses alpha plus beta close parentheses over denominator sin beta end fraction space equals space fraction numerator u space sin space open parentheses alpha space plus space beta space minus space begin display style straight pi over 2 end style close parentheses over denominator sin beta end fraction end style
 
Answered by Thiyagarajan K | 22 May, 2019, 16:29: PM

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