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A particle is projected from a point O with a velocity u at an angle (theta) (upwards) to the horizontal. At a certain point P it moves at right angles to its initial direction. It follows that : (A) OP makes an angle tan -1 (u/2g) to the horizontal (B) the distance of P from O is u/(2gsin(theta)) (C) the time of flight from O to P is u/(gsin(theta)) (D) the velocity of the particle at P is ucot(theta)
A particle is projected from a point O with a velocity u at an angle (theta) (upwards) to the horizontal. At a certain point P it moves at right angles to its initial direction. It follows that :(A) OP makes an angle tan-1(u/2g) to the horizontal(B) the distance of P from O is u/(2gsin(theta))(C) the time of flight from O to P is u/(gsin(theta))(D) the velocity of the particle at P is ucot(theta)

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2 years ago

```							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 cosθ2)  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
```
2 years ago
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