If both are moving in same direction and the submarine is ahead of the warship in both the situations, then the speed of the warship, if the ratio of the speed of warship to that of the submarine is 2 : 1, is:

Let the initial position of the warship be A and initial and final positions of the submarine be P and Q respectively.

$$AP = 50 km$$

Ratio of speed of warship to submarine = $$2 : 1$$

=> $$AB = 2 PQ$$ ---------------Eqn(I)

In $$\triangle$$ APN

=> $$sin 30 = \frac{AN}{AP}$$

=> $$\frac{1}{2} = \frac{AN}{50}$$

=> $$AN = \frac{50}{2} = 25 km$$

=> $$BM = AN = 25 km$$

Similarly, $$PN = 25\sqrt{3} = AR$$

$$\therefore$$ In $$\triangle$$ BMQ

=> $$sin 60 = \frac{BM}{BQ}$$

=> $$\frac{\sqrt{3}}{2} = \frac{25}{BQ}$$

=> $$BQ = \frac{50}{\sqrt{3}} km$$

Similarly,  $$MQ = \frac{25}{\sqrt{3}}$$

Let $$RB = PM = x$$ and using Eqn (I), we get :

=> $$(AR + RB) = 2 (PM + MQ)$$

=> $$(25\sqrt{3} + x) = 2 (x + \frac{25}{\sqrt{3}}$$

=> $$(25\sqrt{3} + x) = 2x + \frac{50}{\sqrt{3}}$$

=> $$2x - x = 25\sqrt{3} - \frac{50}{\sqrt{3}}$$

=> $$x = \frac{25}{\sqrt{3}}$$

$$\therefore AB = 25\sqrt{3} + \frac{25}{\sqrt{3}}$$

=> $$AB = \frac{100}{\sqrt{3}}$$

$$\therefore$$ Speed of warship = $$2 \times \frac{100}{\sqrt{3}} = \frac{200}{\sqrt{3}} km/h$$