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

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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.

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