A cricket ball thrown across a field is at heights h₁ and h₂ from the point of projection at times t₁ and t₂ respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is [WB JEE 2014]

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$(\frac{h_{1} t_{2}^{2} - h_{2} t_{1}^{2}}{h_{1} t_{1} - h_{2} t_{2}})$

$(\frac{h_{1} t_{1}^{2} - h_{2} t_{2}^{2}}{h_{1} t_{2} - h_{2} t_{1}})$

$(\frac{h_{1} t_{2}^{2} - h_{2} t_{1}^{2}}{h_{1} t_{2} - h_{2} t_{1}})$

None of these

answer is C.

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Detailed Solution

For vertical motion, $h_{1} = u \sin θ t_{1} - \frac{1}{2} g t_{1}^{2} (For h_{1})$

$\Rightarrow t_{1} = \frac{h_{1} + \frac{1}{2} g t_{1}^{2}}{u \sin θ}$ ………(i)

$\Rightarrow h_{2} = u \sin θ t_{2} - \frac{1}{2} g t_{2}^{2} (For h_{2})$

$\Rightarrow t_{2} = \frac{h_{2} + \frac{1}{2} g t_{2}^{2}}{u \sin θ}$ ………(ii)

On dividing Eq. (i) by Eq. (ii), we get

$\frac{t_{1}}{t_{2}} = \frac{(h_{1} + \frac{1}{2} g t_{1}^{2}) / u \sin θ}{(h_{2} + \frac{1}{2} g t_{2}^{2}) / u \sin θ}$

$\Rightarrow h_{1} t_{2} - h_{2} t_{1} = \frac{g}{2} (t_{1} t_{2}^{2} - t_{1}^{2} t_{2})$

The time of flight of the ball,

$T = \frac{2 u \sin θ}{g} = \frac{2}{g} (u \sin θ)$

$= \frac{2}{g} (\frac{h_{1} + 1 / 2 g t_{1}^{2}}{t_{1}})$ [From Eq. (i)]

$= \frac{2}{t_{1}} (\frac{h_{1}}{g} + \frac{t_{1}^{2}}{2})$

$= \frac{h_{1}}{t_{1}} \times \frac{2}{g} + t_{1} = \frac{h_{1}}{t_{1}} \times (\frac{t_{1} t_{2}^{2} - t_{1}^{2} t_{2}}{h_{1} t_{2} - h_{2} t_{1}}) + t_{1}$

$= \frac{h_{1} t_{1} t_{2}^{2} - h t_{1}^{2} t_{2} + h t_{1}^{2} t_{2} - h_{2} t_{1}^{3}}{t_{1} (h_{1} t_{2} - h_{2} t_{1})} = (\frac{h_{1} t_{2}^{2} - h_{2} t_{1}^{2}}{h_{1} t_{2} - h_{2} t_{1}})$