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A projectile can have the same range $\mathrm{R}$ for two angles of projection. If $t_{1}$ and $t_{2}$ be the times of flight in two cases, then what is the product of two times of flight?
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The correct answer is:
$\mathrm{t}_{1} \mathrm{t}_{2} \propto \mathrm{R}$
$\mathrm{t}_{1}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}$ and
$\mathrm{t}_{2}=\frac{2 \mathrm{u} \sin (90-\theta)}{\mathrm{g}}=\frac{2 \mathrm{u} \cos \theta}{\mathrm{g}}$
$\begin{aligned} \therefore \mathrm{t}_{1} \mathrm{t}_{2} &=\frac{4 \mathrm{u}^{2} \cos \theta \sin \theta}{\mathrm{g}^{2}}=\frac{2}{\mathrm{~g}}\left[\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\right] \\ &=\frac{2}{\mathrm{~g}} \mathrm{R}, \end{aligned}$
where $\mathrm{R}$ is the range.
Hence $t_{1} t_{2} \propto R$
$\mathrm{t}_{2}=\frac{2 \mathrm{u} \sin (90-\theta)}{\mathrm{g}}=\frac{2 \mathrm{u} \cos \theta}{\mathrm{g}}$
$\begin{aligned} \therefore \mathrm{t}_{1} \mathrm{t}_{2} &=\frac{4 \mathrm{u}^{2} \cos \theta \sin \theta}{\mathrm{g}^{2}}=\frac{2}{\mathrm{~g}}\left[\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\right] \\ &=\frac{2}{\mathrm{~g}} \mathrm{R}, \end{aligned}$
where $\mathrm{R}$ is the range.
Hence $t_{1} t_{2} \propto R$
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