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# Two bodies are projected from the same point with the same speed but in different directions so as to have the same range. The ratio of their times of flight are .

$\begin{array}{1 1}(A)\;1:1\\(B)\;1:\cos \theta \\(C)\;1: \sin \theta\\(D)\;1: \cot \theta \end{array}$

For the same range for the same speed projected if $\theta_1= \theta$ then
$\theta_2 =(90^{\circ}- \theta)$
$t_1=\large\frac{ 2 v_0 \sin \theta}{g}$ and
$t_2 = \large\frac{2 v_0 \cos \theta}{g}$
$\large\frac{t_2}{t_1} =\frac{ \cos \theta}{\sin \theta}$
$\qquad= \cot \theta$
Hence D is the correct answer.