This is war!

We first find the area in which the missile can reach. The equation of trajectory for the missile is

$y = x \tan \theta - \dfrac {\frac {1}{2} g x^2}{u^2} \sec^2 \theta$

For maximum $y$ for a given $x$ -

$\dfrac {\mathrm{d} y}{\mathrm{d} (\tan \theta)} = x - \dfrac {\frac {1}{2} g x^2}{u^2} (2 \tan \theta) = 0$

$\Rightarrow \tan \theta = \dfrac {u^2}{gx}$

Putting in the first equation, we have

$y_{max} = x \dfrac {u^2}{gx} - \dfrac {\frac {1}{2} g x^2}{u^2} \left[1 + \dfrac {u^4}{g^2 x^2} \right] = \dfrac {u^2}{2g} - \dfrac {\frac {1}{2} g x^2}{u^2}$

Therefore, the missile can hit an area given by

$y \leq \dfrac {u^2}{2g} - \dfrac {\frac {1}{2} g x^2}{u^2}$

Given the problem $y = 250m, u = 100m/s, g = 10m/s^2$ . Putting these values we get,

$\dfrac {x^2}{2000} \leq 250 \Rightarrow -500 \sqrt{2} \leq x \leq 500 \sqrt{2}$

Therefore, the plane is in danger for a period of $\dfrac {1000 \sqrt{2}}{500} = 2 \sqrt{2}$ seconds.

We now that the equation of the trajectory of the missile is given by

$y=x \tan\theta-\dfrac{g x^2}{2 v_0^2\cos^2\theta}$

Given that the plane is in danger if the missile reaches $y=250$ m, we can find a function $x(\theta)$ such that the missile reaches such height. If we derivate $x(\theta)$ with respect to $\theta$ and equalise to cero, we can find $\theta$ such that $x$ is maximum. This value times $2$ would be the range in which the fighter plane will be in danger.

So, substituting $y=250$ , $v_0=100$ m/s and $g=10$ m/s $^2$ , we get

$250=x \tan\theta-\dfrac{10 x^2}{2 (10^4)\cos^2\theta}$

or better

$x^2-1000\sin2\theta+5(10^5)\cos^2\theta=0$ .

Solving for $x$ we get

$x(\theta)=500(\sin2\theta\pm\sqrt{\sin^22\theta-\cos^2\theta})$ .

Since we want the maximum value of $x$ , we choose the $+$ sign, and derivate equalising to cero to obtain $\theta =2 \tan ^{-1}\left(\sqrt{2-\sqrt{3}}\right)=54.74$ degrees, which gives $x_{max}$ to be $500\sqrt{2}$ m.

Hence, as said, $1000\sqrt{2}$ m would be the range in which the fighter plane will be in danger, and therefore $2\sqrt{2}=2.828$ seconds would be the duration of time for which the plane is in danger of being hit by the missile.