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(C) For maximum range,the angle of projection is $	heta = 45^{\circ}$. The range $R = \frac{u^2 \sin(2	heta)}{g} = \frac{u^2}{g} = 10^5 \, m$. Thus,

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$500\sqrt{2} \, m/s$

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(C) For maximum range,the angle of projection is $	heta = 45^{\circ}$. The range $R = \frac{u^2 \sin(2	heta)}{g} = \frac{u^2}{g} = 10^5 \, m$. Thus,$u^2 = gR = 10 	imes 10^5 = 10^6$,so $u = 1000 \, m/s$.The horizontal component of velocity is $v_x = u \cos(45^{\circ}) = 1000 	imes \frac{1}{\sqrt{2}} = 500\sqrt{2} \, m/s$.At the maximum height,the vertical component is $0$. At the halfway point of the range,the missile is at its maximum height. Therefore,the speed of the missile when detected is equal to its horizontal component,which is $500\sqrt{2} \, m/s$.

$A$ 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 $g = 10 \, m/s^2$). What was the speed of the missile when it was detected?

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