Whose side are you on Officer?

Solve the simultaneous equations:

$1670\cos(60)t=1670\cos(45)(t-t')\\ 1670\sin(60)t-\frac { 1 }{ 2 } g{ t }^{ 2 }=1670\sin(45)(t-t')-\frac { 1 }{ 2 } g(t-t')$

For both the projectiles the y is the same.
So xtan(60) - 0.5gx^2/v^2cos^2(60) = xtan(45) - 0.5gx^2/v^2cos^2(45). Now solve for x. t1 = x/vcos(60) and t2 = /vcos(45) required time is t1 - t2

The parametric equations of the first projectile are: $x(t_1)=835t_1$ $y(t_1)=835\sqrt{3}t_1-5t_1^2$

And for the second projectile: $x(t_2)=835\sqrt{2}t_2$ $y(t_2)=835\sqrt{2}t_2-5t_2^2$

We need that the coordinates of both projectiles to be the same ( $x(t_1)=x(t_2)$ and $y(t_2)=y(t_2)$ ). So, make the following equation system: $835t_1=835\sqrt{2}t_2$ $835\sqrt{3}t_1-5t_1^2=835\sqrt{2}t_2-5t_2^2$

It's easy to see from the first equation that $t_1=t_2\sqrt{2}$ . Replace that value on the second equation and solve: $835\sqrt{3}(t_2\sqrt{2})-5(t_2\sqrt{2})^2=835\sqrt{2}t_2-5t_2^2$ $t_2=167\sqrt{6}-167\sqrt{2}$

Now, obtain $t_1$ : $t_1=(167\sqrt{6}-167\sqrt{2})(\sqrt{2})$ $t_1=334\sqrt{3}-334$

And, our answer, by logic, will be $t_1-t_2$ , so: $t_1-t_2=167\sqrt{2}+334\sqrt{3}-167\sqrt{6}-334\approx\boxed{71.61}$

Given a diagram showing the two projectiles fired at angles $\theta$ and $\phi$ Projectile The first projectile was fired at t = 0 and let us assume that the second projectile was fired at t = $\Delta$ t time after the first one.
Since the second projectile must destroy the first one in its path there must be a common point in the trajectories of both projectile
Let us assume that the projectile collide at a point (R,h) in their path. So the point (R,h) must satisfy the equations of projectile.
In Vertical Components
Vertical component Where v is the inital velocity of both the projectiles and since the second one is fired after $\Delta$ t time after its time to reach h will be t - $\Delta$ t
In Horizontal component
Horizontal component

Solving for $\Delta$ t we get
Final Expression
Substituting the value of V, $\theta$ , $\phi$ , and g we get Solution

Note. This problem is very, very far from reality because .... lacks air. :)