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(D) For maximum range,the angle of projection is $	heta = 45^{\circ}$.The range $R$ is given by $R = \frac{u^2 \sin(2	heta)}{g}$. Since $	heta = 45

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$\frac{100}{\sqrt{2}} \, s$

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(D) For maximum range,the angle of projection is $	heta = 45^{\circ}$.The range $R$ is given by $R = \frac{u^2 \sin(2	heta)}{g}$. Since $	heta = 45^{\circ}$,$R = \frac{u^2}{g}$.Given $R = 100 \, km = 10^5 \, m$ and $g = 10 \, m/s^2$,we have $u^2 = gR = 10 	imes 10^5 = 10^6 \, m^2/s^2$,so $u = 1000 \, m/s$.The total time of flight is $T = \frac{2u \sin 	heta}{g} = \frac{2 	imes 1000 	imes \sin(45^{\circ})}{10} = 200 	imes \frac{1}{\sqrt{2}} = 100\sqrt{2} \, s$.The missile is detected at its half-way point,which corresponds to the peak of its trajectory (since the range is symmetric for $45^{\circ}$ projection).The time taken to reach the peak is $t_{peak} = \frac{T}{2} = \frac{100\sqrt{2}}{2} = 50\sqrt{2} \, s$.The remaining time until impact is $T - t_{peak} = 100\sqrt{2} - 50\sqrt{2} = 50\sqrt{2} \, s$.Wait,re-evaluating the detection point: The half-way point in terms of horizontal distance is the peak. The time remaining is the time from the peak to the ground,which is $t = \frac{u \sin 	heta}{g} = \frac{1000 	imes (1/\sqrt{2})}{10} = \frac{100}{\sqrt{2}} \, s$.

$A$ missile is fired for maximum range at your town from a place $100 \, km$ away from you. If the missile is first detected at its half-way point,how much warning time will you have? (Take $g = 10 \, m/s^2$)

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