# A particle is projected from a point A with a velocity u at an angle θ to the horizontal.

+1 vote
2.6k views
in JEE

A particle is projected from a point A with a velocity u at an angle θ to the horizontal. At a certain point B, it moves at right angle to its initial direction.

Can anyone provide solution for all the questions in the image

+1 vote
by (21.1k points)

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

by (45 points)
wud save too much time
+1 vote
by (45 points)

Feel the vectors!!!