Sarthaks Test
+1 vote
in JEE by (182 points)

A particle is projected from a point A with a velocity u at an angle θ to the horizontal. At a certain point B, it moves at right angle to its initial direction.

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2 Answers

+1 vote
by (21.1k points)

The question is not very clear.  

1) I understand that v is the component of instantaneous velocity of the projectile in the direction perpendicular to the initial direction of projection.
then:  v = (u cosθ) sinθ - (u sinθ - g t) cosθ = g t cosθ

2)  vx = u cosθ               vy = u sinθ - g t
     let the direction of v = Φ.
     tanΦ = vy/vx = (u sinθ - g t) / (u cosθ)
     given vectors u and u are perpendicular.  so tanΦ = - cotθ
        (u sin θ - g t) sinθ = - u cos² θ 
        =>  u = g t sin θ

     Now  vx = u cosθ = g t sinθ cosθ
              vy = gt sin² θ - g t = - g t cos²θ

      so  v = √(vx² + vy²) = gt cosθ  = u cot θ

by (45 points)
use vector method instead
wud save too much time
+1 vote
by (45 points)

Feel the vectors!!!

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