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 A particle is projected under gravity with a velocity u in a direction making an  angle a with the horizon. Show that the amount of deviation D in direction of  motion of the particle is given by : tan ( cos ) / ( sin )
Asked by dhandas060104 | 24 Oct, 2020, 21:29: PM
answered-by-expert Expert Answer
Let uo be initial projection velocity at an angle α with horizontal
 
At any instant t, vertical component of velocity = ( uo sinα )- gt
 
At any instant t, horizontal component of velocity = uo cosα
 
Angle β made by direction of motion with horizontal is determined as
 
tanβ = [ ( uo sinα )- gt ] / ( uo cosα )  = tanα - [ (gt) / (uo cosα ) ] ...................(1)
 
From above equation-(1) , we get , tanα - tanβ = [ (gt) / (uo cosα ) ]  ....................(2)
 
also we get from eqn.(1) , tanα tanβ = tan2α - [ ( g t sinα ) / ( uo cos2α ) ]  .................(3)
 
Angle of deviation D in direction of motion,  D = ( α - β)
 
tanD = tan(α - β ) = [ tanα - tanβ ] / [ 1 - (tanα tanβ) ] .............. (4)
 
Using eqn.(2) and eqn.(3) , eqn.(4) is rewritten after simplification as
 
begin mathsize 14px style tan D space equals space fraction numerator g t space cos alpha over denominator u subscript o cos 2 alpha space minus space g t space sin alpha space end fraction end style
 
 
Answered by Thiyagarajan K | 25 Oct, 2020, 10:07: AM

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