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Question

(D) The cannon shell can reach any point $(x, y)$ that satisfies the equation of the safety envelope (bounding parabola). The equation for the maximum

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$2\sqrt{2}\ s$

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(D) The cannon shell can reach any point $(x, y)$ that satisfies the equation of the safety envelope (bounding parabola). The equation for the maximum height $y$ reachable at a horizontal distance $x$ is given by $y = \frac{u^2}{2g} - \frac{gx^2}{2u^2}$.Given $y = 250\ m$,$u = 100\ m/s$,and $g = 10\ m/s^2$,we substitute these values:$250 = \frac{100^2}{2(10)} - \frac{10x^2}{2(100^2)}$$250 = \frac{10000}{20} - \frac{10x^2}{20000}$$250 = 500 - \frac{x^2}{2000}$$\frac{x^2}{2000} = 250$$x^2 = 500,000$$x = \sqrt{500,000} = 500\sqrt{2}\ m$.The plane is in danger while it travels from $x = -500\sqrt{2}\ m$ to $x = +500\sqrt{2}\ m$,covering a total distance of $1000\sqrt{2}\ m$.The time duration is $t = \frac{	ext{distance}}{	ext{velocity}} = \frac{1000\sqrt{2}}{500} = 2\sqrt{2}\ s$.

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