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

$y^{\prime}=x \tan \theta-\frac{\frac{1}{2} g x^{2}}{u^{2}} \sec ^{2} \theta \cdots \cdots(1)$

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

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

$\Rightarrow \tan \theta=\frac{u^{2}}{g x}$

Putting in equation $(1)$

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

The cannon shell can hit an area given by

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

Given in the problem$\mathrm{y}=250 \mathrm{m}$

$\mathrm{u}=100 \mathrm{m} / \mathrm{s}, \mathrm{g}=10 \mathrm{m} / \mathrm{s}$

Putting these value we get,

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

Plane in danger for a period of 

$\frac{{1000\sqrt 2 }}{{500}} = 2\sqrt 2 \,\sec $