JEE Class main Answered
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, 09:29: PM
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
Answered by Thiyagarajan K | 25 Oct, 2020, 10:07: AM
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