Question

а s(b) A body is projected with a velocity of loma' obliquely . Find the angle of projection of its right. range is twice its masimum​

Answer 1

Answer:

At time t=0 (when body is projected),

Horizontal velocity, ux=ucosα

Vertical velocity, uy=usinα

Velocity vector, u=ucosαi^+usinαj^

At any time t;

Horizontal velocity, vx=ucosα

Vertical velocity, vy=usinα−gt

Velocity vector,  v=ucosαi^+(usinα−gt)^j

When two vectors are perpendicular, their dot product is zero.

u.v=0

This implies (ucosα)2+(usinα)2−(usinα×gt)=0

t=gsinαu

∴vy=usinα−g(gsinαu)=u(sinα−cosecα)

 

Velocity magnitude, ∣v∣=((ucosα)2+(u(sinα−