Question

32.A projectile is fired with a speed 'U' at an angle 'theeta' with the horizontal It's speed when it's direction of motion makes an angle 'alpha' with the horizontal is

Answer 1

horizontal component of velocity = vx = u cos θ
vertical component of velocity = vy = u sinθ - g t

angle of velocity with horizontal = α =>
     tan α = vy/vx = (u sinθ - g t) / (u cos θ)
     t = u (sinθ - cos θ tan α ) / g

     vy = u cosθ tan α
     vx = u cosθ

So the speed of the projectile = √[ vx² + vy² ] = u cosθ secα .