a particle is projected from point A with velocity u at an angle a with horizontal,. In its parabolic path, at point P the particle is moving at right angles to its initial direction of projrction . Its velocity at P is

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Solution

$L e t t h e a n g l e m a d e b y p r o j e c t i l e w i t h t h e h o r i z o n t a l b e θ . A t p o i n t P t h e a n g l e m a d e b y t h e p r o j e c t i l e w i t h t h e d i r e c t i o n o f i n i t i a l p r o j e c t i o n i s 90 °, s o t h e a n g l e m a d e w i t h t h e h o r i z o n t a l i s (90 ° - θ) . T h e r e f o r e u \cos θ = v \cos (90 ° - θ) o r u c o s θ = v \sin θ o r v = u c o t θ S o v e l o c i t y a t p o i n t P i s u c o t θ .$

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a particle is projected from point A with velocity u at an angle a with horizontal,. In its parabolic path, at point P the particle is moving at right angles to its initial direction of projrction . Its velocity at P is

Let the angle made by projectile with the horizontal be θ.At point P the angle made by the projectile with the direction of initial projection is 90°, so the angle made with the horizontal is 90°-θ.Therefore u cosθ=v cos 90°-θor u cosθ=v sinθor v= u cotθSo velocity at point P is u cotθ.