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Question

(b) For same range angle of projection should be $	heta$ and $90-	heta$

So, time of flights ${t_1} = \frac{{2u\sin 	heta }}{g}$ and

${t_2} =

Vedclass · Accepted Answer

${t_1}{t_2} \propto \,R$

Vedclass · Answer

(b) For same range angle of projection should be $	heta$ and $90-	heta$

So, time of flights ${t_1} = \frac{{2u\sin 	heta }}{g}$ and

${t_2} = \frac{{2u\sin (90 - 	heta )}}{g} = \frac{{2u\cos 	heta }}{g}$

By multiplying$ = {t_1}{t_2} = \frac{{4{u^2}\sin 	heta \cos 	heta }}{{{g^2}}}$

${t_1}{t_2} = \frac{2}{g}\frac{{({u^2}\sin 2	heta )}}{g} = \frac{{2R}}{g}$

$&rArr;$ ${t_1}{t_2} \propto R$

For a given velocity, a projectile has the same range $R$ for two angles of projection if $t_1$ and $t_2$ are the times of flight in the two cases then

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