Question

A particle is projected with velocity u making an angle theta with horizontal at any instant its velocity v is a right angle to the initial velocity u then velocity is

Answer 1

Hiii........

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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

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hope it will help u.......

Answer 2

Hiii........

========================================

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

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hope it will help u.......