TWO particles are thrown simultaneously with a velocity of 30m/s and one is thrown vertically upward and another at 45 degree with respect to the horizon . find out the distance between them at time = 1.5 second .

here we can use the concept of relative velocity...let the first particle be at rest ...so it remains at origin ... but we have to add -30m/s to its velocity for it to come at rest.. hence we must add the same to the vertical component of the velocity of the second particle.. hence the vertical component of initial velocity becomes 30sin45-30... now there is avertical acceleration of g downwards.. hence y= ut +1/2at² = -8.79t-5t² =-13.185-11.25=-24.435.... now horizontal velocity is constant... hence x=ut=30cos45*t=2.115... now the distance between origin and the point would be.. v(x²+y²) =v(173.844+4.473)=v178.317=13.35m