Projectile again

Since horizontal velocity for a projectile is only possible at the maximum height, both are at equal height with vertical velocity zero. So after t sec. both will have the same vertical velocity=gt m/s, but horizontal constant velocity 3 and -4 m/s. So the angles will be $\phi_3~~ and~~ \phi_4.~~~Tan\phi_3=-Cotan \phi_4.\\\implies \dfrac{gt} 3=-\dfrac{-4}{gt},~~~~\therefore ~g*t^2=12.\\So~distance~ between ~them=\{3-(-4)\}*\sqrt{\dfrac{12} g } =~~~~~~~ \color{#D61F06}{2.425} m$

Since the velocities are perpendicular: $v_{1}\cdot v_{2} = 0$

That is: $-12+ g^{2}t^{2} = 0$

Solve for t: $t=(\frac{12}{g^{2}})^{\frac{1}{2}}$

Then: 3t + 4t = 2.425

Since they are at equal heights, the horizontal distance between them is the correct answer.

Let the moment in question be t. Then equal vertical components of the particles become gt. As the vectors of the particles velocities v1 and v2 become perpendicular, we get the equation gt/v1 = v2/gt. Which means that t = sqrt (v1v2)/g. In the reference frame falling with the particles with accelaration g the distance L between them becomes L = (v1+v2) sqrt (v1v2)/g = 2.425 m.