Trajactories

First, we must record our constants:

${\theta_1}=60º$ , ${\theta_2}=0º$ , ${v_0}=5\sqrt{3}\ m/s$ and ${v_0}=10\ m/s^{2}$ .

It's important to note that the initial time is diferent to both movements, since they begin at different instants. From the exposed, we have:

${t'_0}={t_0}+\Delta{t}$

We can fix ${t_0}=0$ and then assemble the parametric equations of the two movements as follows:

${x_1}(t)={v_0}.cos\ {\theta_1}.(t-{t_0})$

${y_1}(t)={v_0}.sen\ {\theta_1}.(t-{t_0})-{\frac{g}{2}}.(t-{t_0})^{2}$

${x_2}(t)={v_0}.cos\ {\theta_2}.(t-{t'_0})$

${y_2}(t)={v_0}.sen\ {\theta_2}.(t-{t'_0})-{\frac{g}{2}}.(t-{t'_0})^{2}$

At the instant the two shells collide, both vertical and horizontal position are equal. So, we have:

${x_1(t)}={x_2}(t)$

${v_0}.cos\ {\theta_1}.t={v_0}.cos\ {\theta_2}.(t-\Delta{t})$

$5\sqrt{3}.cos\ 60º.t=5\sqrt{3}.cos\ 0º.(t-\Delta{t})$

$cos\ 60º.t=cos\ 0º.(t-\Delta{t})$

$\frac{1}{2}.t=1.(t-\Delta{t})$

$\Delta{t}=\displaystyle {\frac{t}{2}}$

${y_1(t)}={y_2}(t)$

${v_0}.sen\ {\theta_1}.t-\displaystyle{\frac{g}{2}}.t^{2}={v_0}.sen\ {\theta_2}.(t-\Delta{t})-{\frac{g}{2}}.(t-\Delta{t})^{2}$

${v_0}.sen\ {\theta_1}.t-{v_0}.sen\ {\theta_2}.(t-\Delta{t})=\displaystyle {\frac{g}{2}}.t^{2}-{\frac{g}{2}}.(t-\Delta{t})^{2}$

${v_0}.[sen\ {\theta_1}.t-sen\ {\theta_2}.(t-\Delta{t})]=\displaystyle {\frac{g}{2}}.[t^{2}-(t-\Delta{t})^{2}]$

${v_0}.[sen\ {\theta_1}.t-sen\ {\theta_2}.(t-\displaystyle {\frac{t}{2}})]={\frac{g}{2}}.[t^{2}-(t-{\frac{t}{2}})^{2}]$

${v_0}.[sen\ {\theta_1}.t-sen\ {\theta_2}.(t-{\frac{t}{2}})]=\displaystyle{\frac{3gt^{2}}{8}}$

$5\sqrt{3}.\displaystyle \frac{\sqrt{3}}{2}.t-0.(t-{\frac{t}{2}})=\displaystyle{\frac{3gt^{2}}{8}}$

$\displaystyle \frac{15}{2}.t=\displaystyle{\frac{3.10}{8}.t^{2}}$

And finally:

$t=2 \rightarrow \Delta{t}=\displaystyle {\frac{2}{2}}=\boxed{1\ s}$

First of all for collision time of flight of firstly fired bullet must be greater than the time if flight. Suppose they collide $t +\Delta$ time after the first bullet has been fired or $t$ t time after the first bullet has been fired. $i.e. t +\Delta \leq \frac{5√3sin(60°) + \sqrt{(5√3sin(60°))^2 + 2g\times 10}}{g}$ $t+\Delta \leq 2.35078$ For collision, distance travelled in y dir and distance travelled in x dir will be same for both the bullets. $\Rightarrow 5√3cos(60°) (t+\Delta) = 5√3 t \space \space$ Here we get, $t= \Delta. (i)$ And $5√3 sin(60°)(t+ \Delta) - g(t+\Delta)^2/2 = -gt^2/2 \space \space (ii)$ From (i) &(ii) $15 = 15t \Rightarrow \boxed{t=1}$

You really need to specify that the times are from firings to the moment when both shells meet at the same point in space at the same time. That's a detail that's left out, I think.