A fighter jet is flying horizontally at a certain altitude with a speed of $$200$$ m s$$^{-1}$$. When it passes directly overhead an anti-aircraft gun, a bullet is fired from the gun, at an angle $$\theta$$ with the horizontal, to hit the jet. If the bullet speed is $$400$$ m s$$^{-1}$$, the value of $$\theta$$ will be ______ °.

A fighter jet is flying horizontally at 200 m/s. When it passes directly overhead an anti-aircraft gun, a bullet is fired at 400 m/s at angle $$\theta$$ with the horizontal to hit the jet. We need to find $$\theta$$.

We set up a coordinate system by placing the gun at the origin. At the moment of firing ($$t = 0$$), the jet is directly overhead at position $$(0, h)$$, where $$h$$ is the altitude, and it moves in the positive x-direction with speed $$v_{jet} = 200\text{ m/s}$$.

At time $$t$$, the jet’s position is therefore $$x_{jet} = 200t,\quad y_{jet} = h\,.$$

The bullet is fired with speed 400 m/s at angle $$\theta$$ above the horizontal, giving horizontal and vertical components $$v_{bx} = 400\cos\theta,\quad v_{by} = 400\sin\theta\,.$$ At time $$t$$, its coordinates are $$x_{bullet} = 400\cos\theta \cdot t,\quad y_{bullet} = 400\sin\theta \cdot t - \tfrac12 gt^2\,.$$

For the bullet to hit the jet at some time $$t=T$$, both x- and y-coordinates must coincide. Equating the horizontal positions gives $$400\cos\theta\;T = 200T\,. $$ Since $$T\neq0$$, division by $$T$$ yields $$400\cos\theta = 200\quad\Rightarrow\quad\cos\theta = \tfrac12\,. $$

Matching the vertical coordinates requires $$400\sin\theta\;T - \tfrac12 gT^2 = h\,, $$ which determines the time $$T$$ at which the bullet reaches altitude $$h$$. As long as a positive $$T$$ exists (which it does for any reachable altitude), this condition is satisfied once $$\theta$$ satisfies the horizontal equation. Therefore, the firing angle is determined solely by $$\cos\theta = \tfrac12$$.

Solving $$\cos\theta = \tfrac12$$ gives $$\theta = \cos^{-1}\bigl(\tfrac12\bigr) = 60^\circ\,. $$ With $$\theta = 60^\circ$$, the vertical component is $$v_{by} = 400\sin60^\circ = 400\times\tfrac{\sqrt3}{2} = 200\sqrt3\approx 346.4\text{ m/s},$$ and the vertical motion equation becomes $$200\sqrt3\,T - 5T^2 = h\,, $$ which indeed has a positive solution for any feasible altitude $$h$$, confirming that the bullet reaches the jet.

Therefore, the required firing angle is 60°.