Two guns situated at the top of a hill of height 10m, fire one shot each with the same speed of $5\sqrt{3}m{s}^{-1}$  at some interval of time. One gun fires horizontally and other fires upwards at an angle of ${60}^0$  with the horizontal. The shots collide in air at the point P. Taking the origin of the coordinate system at the foot of the hill right below the muzzle, trajectories in  $x-y$ plane and $g=10m{s}^{-2}$, then

Let the second shell reach the point  $P(x,y)$ at time  $t_2$ and the first shell be fired at a time
$(t_2-\Delta t)-t_1$

For shell 1
$$\begin{gathered} x=\left(5\sqrt{3}\right)t_1 \\ =(10-y)=\frac{1}{2}g{t}_1^2 \end{gathered}$$

For shell 2
$$\begin{gathered} x=\left(5\sqrt{3}\cos 60\right)t_2 \\ =(10-y)=\left(5\sqrt{3}\sin 60\right)t_2=\frac{1}{2}g{t}_2^2 \end{gathered}$$

Equating (1) and (2) we get
$t_2=2t_1$
Equating(2) and (3) we get
$$\begin{gathered} -\frac{1}{2}g{t}_1^2=\left(5\sqrt{3}\sin 60\right)t_2-\frac{1}{2}g{t}_2^2 \\ \Rightarrow -\frac{1}{2}g{t}_1^2=\frac{15}{2}\left(2t_1\right)-\frac{1}{2}g{\left(2t_1\right)}^2 \\ \Rightarrow -\frac{1}{2}g{t}_1^2=15t_1-2g{t}_1^2 \\ -5t_1^2-15t_1=20t_1^2 \\ \Rightarrow 5t_1^2=15t_1 \\ \Rightarrow 5t_1^2-15t_1=0 \\ \Rightarrow 15t_1(t_1-1)=0 \\ \Rightarrow t_1=1s \\ \Rightarrow t_2=2s,i.e\Delta t=1s \\ \Rightarrow x=5\sqrt{3}mandy=5m \end{gathered}$$