Question

A particle is projected with a speed u at an angle θ with the horizontal. In its parabolic path at a point, particle is moving at right angle to initial direction of projection. What will be it's velocity at that point?

Answer 1

hey mate
here's the solution

Answer 2

Let the initial velocity of object  be  u which makes  \theta from  the  horizontal line.

horizontal component  of  vector  u = ucosθ

Let  P a  point  where velocity of object  is  v and  the direction of  velocity will be  tangent  on parabolic  path at  point  P.

According  to question, velocity vector  v is perpendicular  to velocity vector  u

Therefore  angle  between u and  v  = 90°

Now, as velocity vector makes  and  angle  of  \theta  from horizontal line , velocity velocity  vecto rv will make  angle of  90-\theta from horizontal line.  

Now, horizontal component  of vetor  v,= vcos(90-θ) = vsinθ  

We know that  , during  the  projectile  motion of  an object  horizontal component  of  velocity  remains same while  vertical component  changes due  to force of  gravity in opposite  direction.

Therefore,

horizontal component  at  initial point  and  horizontal component  at  point  P will be  same.

ucosθ = vsinθ

ucosθ/sinθ = v

v = ucotθ