Question

two equap velocities have a resultant equal to 3/2 times the value of either velocity.find the angle between them.​

Answer 1

Let R be the resultant of the velocities with an angle Ф between them.

   R² =  (3V/2)² = V² + V² + 2 V V Cos Ф 

     9 V²/4 = 2 V² + 2 V² Cos Ф

   Cos Ф = 1/8 
    Ф = 82.82 deg

and it can be solved also this method resulatnt velocity is given by:R = v12+v22+2v1v2cosθwhere θ is the angle between them.
v1 = v2 = vR = 32v 32v =  v2+v2+2v2cosθ32 = 2+2cosθ94 = 2+2cosθ1+cosθ = 98cosθ = 98-1cosθ = 18θ = cos-1(1/8)

Answer 2

Answer:

let force 1=A, force2=B resultant equal to 3/2A

hereA=B

Explanation:

A^2 +B^2+2ABcostheta=R^2

A^2+A^2+2AAcostheta =(3/2×A)^2

2A^2+2A^2 costheta =9/4×A^2

2A^2(1+costheta) =9/4A^2

1+costheta =9/4A^2×1/2A^2

1+costheta =9/8

costheta=9/8-1

costheta =1/8

theta=1/cos ×1/8

theta=~83