Two particles are projected from the same point on the ground, with same speed simultaneously at angles $α$ and $β$ with the horizontal. They strike the ground at same point after times $t_{1}$ and $t_{2}$ respectively. Then: $(α > β)$

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$\frac{t_{1}}{t_{2}} = \tan α$

$α - β = 90^{0}$

$\frac{t_{1}}{\sin α} = \frac{t_{2}}{\sin β}$

$\frac{t_{1}}{t_{2}} = \cot β$

answer is B, C, D.

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Detailed Solution

To get same horizontal range, $α + β = 90^{0}$

$α = 90^{0} - β$
$t_{1} = \frac{2 u \sin α}{g}$
$t_{2} = \frac{2 u \sin β}{g} = \frac{2 u}{g} \sin (90 - α) = \frac{2 u \cos α}{g}$
$\frac{t_{1}}{t_{2}} = \tan α & \frac{t_{1}}{\sin α} = \frac{t_{2}}{\sin β} = \frac{2 u}{g}$

$\frac{t_{1}}{t_{2}} = \tan (90 - β) \Rightarrow \frac{t_{1}}{t_{2}} = \cot β$