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A projectile can have the same range 'R' for two angles of projection.If $t_1$ and $t_2$ be the time of flights in the two cases,then the product of the two time of flights is proportional to

$\begin{array}{1 1}R\\\large\frac{1}{R}\\\large\frac{1}{R^2}\\R^2\end{array} $

1 Answer

Answer : R
$t_1=\large\frac{2v\sin \theta}{g}$
$t_2=\large\frac{2v\sin (90^{\large\circ}-\theta)}{g}=\frac{2v\cos\theta}{g}$
$\Rightarrow \large\frac{2}{g}\bigg[\frac{v^2(2\sin \theta\cos\theta)}{g}\bigg]=\frac{2}{g}\frac{v^2\sin 2\theta}{g}=\frac{2}{g}=$$R$
$\therefore t_1t_2\propto R$
answered Aug 13, 2014 by sreemathi.v

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