A particle is thrown at angle theta from the horizontal with initial velocity u from the ground find the time when velocity will be perpendicular to initial velocity u
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Answer:
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
4.5
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