Question

A particle is projected with speed v at an angle theta to the horizontal on an inclined surface making an angle Lambda to the horizontal . find an expression for its range along its along the inclined surface​

Answer 1

Answer:

Explanation:

c) u cot θ

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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