Two particles projected from the same point with same speed u at angles of projection α and ß strike the horizontal ground at the same point. If h1 and h2 are the maximum heights attained by projectiles, R is the range for both and t1 and t2 are their time of flights respectively then

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Two particles projected from the same point with same speed u at angles of projection α and ß strike the horizontal ground at the same point. If h₁ and h₂ are the maximum heights attained by projectiles, R is the range for both and t₁ and t₂ are their time of flights respectively then

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$\frac{t_{1}}{t_{2}} = \tan α$

$\tan α = \sqrt{h_{1} / h_{2}}$

$R = 4 \sqrt{h_{1} h_{2}}$

$α + β = \frac{π}{2}$

answer is A, B, C, D.

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Detailed Solution

Horizontal ranges are same for angle α and ß =(90 - α ) for same projection speed
$\begin{array}{l} h_{2} = \frac{u^{2}}{2 g} \sin^{2} α and h_{2} = \frac{u^{2}}{2 g} \cos^{2} α \\ R = \frac{u^{2}}{g} 2 \sin αcos α \\ R = \frac{u^{2}}{g} \times 2 \frac{\sqrt{2 {gh}_{1}}}{u} \frac{\sqrt{2 {gh}_{2}}}{u} \\ R = 4 \sqrt{h_{1} h_{2}} \\ \frac{t_{1}}{t_{2}} = \frac{2 usin α / g}{2 usin (90 - α) / g} = \tan α \end{array}$