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
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
${t_1}{t_2} \propto \,{R^2}$
${t_1}{t_2} \propto \,R$
${t_1}{t_2} \propto \,\frac{1}{R}$
${t_1}{t_2} \propto \,\frac{1}{{{R^2}}}$
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
For same range angle of projection should be $\theta$ and $90-\theta$
So, time of flights ${t_1} = \frac{{2u\sin \theta }}{g}$ and
${t_2} = \frac{{2u\sin (90 - \theta )}}{g} = \frac{{2u\cos \theta }}{g}$
By multiplying$ = {t_1}{t_2} = \frac{{4{u^2}\sin \theta \cos \theta }}{{{g^2}}}$
${t_1}{t_2} = \frac{2}{g}\frac{{({u^2}\sin 2\theta )}}{g} = \frac{{2R}}{g}$
$⇒$ ${t_1}{t_2} \propto R$
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