A cricket ball thrown across a field is at heights $h_{1}$ and $h_{2}$ from the point of projection at times $t_{1}$ and $t_{2}$ 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:

\frac{h_{1} t_{2}^{2} - h_{2} t_{1}^{2}}{h_{1} t_{2} - h_{2} t_{1}}

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\frac{h_{1} t_{1}^{2} + h_{2} t_{2}^{2}}{h_{2} t_{1} + h_{1} t_{2}}

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\frac{h_{1} t_{2}^{2} + h_{2} t_{1}^{2}}{h_{1} t_{2} + h_{2} t_{1}}

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\frac{h_{1} t_{1}^{2} - h_{2} t_{2}^{2}}{h_{1} t_{1} - h_{2} t_{2}}

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Solution

The correct option is A $\frac{h_{1} t_{2}^{2} - h_{2} t_{1}^{2}}{h_{1} t_{2} - h_{2} t_{1}}$
For vertical movement of the ball
$h_{1} = u sin θ t_{1} - \frac{1}{2} g t_{1}^{2}$
or $t_{1} = \frac{h_{1} + \frac{1}{2} g t_{1}^{2}}{u sin θ}$ .....(i)
Similarly,
$h_{2} = u sin θ t_{2} - \frac{1}{2} g t_{2}^{2}$
or $t_{2} = \frac{h_{2} + \frac{1}{2} g t_{2}^{2}}{u sin θ}$ ......(ii)
Divide equation (i) by equation (ii)
$\frac{t_{1}}{t_{2}} = \frac{\frac{h_{1} + \frac{1}{2} g t_{1}^{2}}{u sin θ}}{\frac{h_{2} + \frac{1}{2} g t_{2}^{2}}{u sin θ}}$
$\Rightarrow h_{1} t_{2} - h_{2} t_{1} = \frac{g}{2} (t_{1} t_{2}^{2} - t_{1}^{2} t_{2})$
The time of flight of the ball
$T = \frac{2 u sin θ}{g} = \frac{2}{g} (u sin θ)$ [from equation (i)]
$= \frac{2}{g} ⎡ ⎢ ⎢ ⎢ ⎣ \frac{h_{1} + \frac{1}{2} g t_{1}^{2}}{t_{1}} ⎤ ⎥ ⎥ ⎥ ⎦ = \frac{2}{t_{1}} [\frac{h_{1}}{g} + \frac{t_{1}^{2}}{2}]$
$= \frac{h_{1}}{t_{1}} \times \frac{2}{g} + t_{1} = \frac{h_{1}}{t_{1}} \times (\frac{t_{1} t_{2}^{2} - t_{1}^{2} t_{2}}{h_{1} t_{2} - h_{2} t_{1}}) + t_{1}$
$= \frac{h_{1} t_{1} t_{2}^{2} - h_{1} t_{1}^{2} t_{2} + h_{1} t_{1}^{2} t_{2} - h_{2} t_{1}^{3}}{t_{1} (h_{1} t_{2} - h_{2} t_{1})}$
$= (\frac{h_{1} t_{2}^{2} - h_{2} t_{1}^{2}}{h_{1} t_{2} - h_{2} t_{1}})$

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Similar questions

When two projectiles are fired from complementary angles having times of flight T_1,T₂and maximum heights H₁ , H₂respectively.

Which of the following is correct?

1) R= gT₁T₂/2

2)R= 4√H₁H₂

3) angle of protection=tan^-1*T₁/T₂

4) angle of projection= tan^-1*H₁/H₂

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