A body is projected at an angle $θ$ with horizontal. Another body is projected with the same velocity at an angle $θ$ with the vertical. The ratio of the time of flights is-

1 : 1

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t a n^{2} θ : 1

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1 : c o t θ

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t a n^{2} 1 : θ

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Solution

The correct option is D $1 : c o t θ$

Let time of first and second projectiles are $t_{1} a n d t_{2}$ respectively.

$t_{1} = \frac{2 u s i n θ}{g} t_{2} = \frac{2 u s i n (90 - θ)}{g} = \frac{2 u c o s θ}{g} \frac{t_{1}}{t_{2}} = \frac{\frac{2 u s i n θ}{g}}{\frac{2 u c o s θ}{g}} = \frac{s i n θ}{c o s θ} = \frac{1}{c o t θ}$

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