Two projectiles are thrown from the same point simultaneously with same velocity 10 ms^-1. One goes straight vertically while other at 60° with the vertical. What will be the distance of separation between the two after 1 second of their throw?

Here for the projectile thrown at 60° with the vertical.
Horizontal velocity = 10 sin 60° = $5\sqrt{3}{\mathrm{ms}}^{-1}$
vertical velocity = 10 cos 60° = 5 ms^-1
Horizontal separation $\mathrm{\Delta x}=5\sqrt{3}\times 1=5\sqrt{3}$
Vertical separation = h₁ - h₂
$\begin{array}{l}\mathrm{\Delta y}=\left(10\mathrm{t}-\frac{1}{2}g{\mathrm{t}}^2\right)-\left[5\mathrm{t}-\frac{1}{2}g{t}^2\right]\\ =(10\times 1-5\times 1)=5\mathrm{m}\end{array}$
$\therefore$displacement $\left|\mathrm{r}\right|=\sqrt{(5\sqrt{3}{)}^2+5^2}=10\mathrm{m}$