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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
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Verified Answer
The correct answer is:
$\frac{h_{1} t_{2}^{2}-h_{2} t_{1}^{2}}{h_{1} t_{2}-h_{2} t_{1}}$
For vertically moment (for $\left.h_{1}\right)$
or
$\begin{array}{l}
h_{1}=u \sin \theta t_{1}-\frac{1}{2} g t_{1}^{2} \\
t_{1}=\frac{h_{1}+\frac{1}{2} g t_{1}^{2}}{u \sin \theta}
\end{array}$
$h_{2}=u \sin \theta t_{2}-\frac{1}{2} g t_{2}^{2}$
or $t_{2}=\frac{h_{2}+\frac{1}{2} g t_{2}^{2}}{u \sin \theta}$
Divide Eq. (i) by Eq. (ia)
$\begin{array}{l}
\frac{t_{1}}{t_{2}}=\frac{h_{1}+\frac{1}{2} g t_{1}^{2} / u \sin \theta}{h_{2}+\frac{1}{2} g t_{2}^{2} / u \sin \theta} \\
\Rightarrow h_{1} t_{2}-h_{2} t_{1}=\frac{g}{2}\left(t_{1} t_{2}^{2}-t_{1}^{2} t_{2}\right)
\end{array}$
The time of flight of the ball $T=\frac{2 u \sin \theta}{g}=\frac{2}{g}(u \sin \theta) \quad$
$=\frac{2}{g}\left[\frac{h_{1}+\frac{1}{2} g t_{1}^{2}}{t_{1}}\right]=\frac{2}{t_{1}}\left[\frac{h_{1}}{g}+\frac{t_{1}^{2}}{2}\right]$
$=\frac{h_{1}}{t_{1}} \times \frac{2}{g}+t_{1}=\frac{h_{1}}{t_{1}} \times\left(\frac{t_{1} t_{2}^{2}-t_{1}^{2} t_{2}}{h_{1} t_{2}-h_{2} t_{1}}\right)+t_{1}$
$=\frac{h_{1} t_{1} t_{2}^{2}-h_{1} t_{1}^{2} t_{2}+h_{1} t_{1}^{2} t_{2}-h_{2} t_{1}^{3}}{t_{1}\left(h_{1} t_{2}-h_{2} t_{1}\right)}$
$=\left(\frac{h_{1} t_{2}^{2}-h_{2} t_{1}^{2}}{h_{1} t_{2}-h_{2} t_{1}}\right)$
or
$\begin{array}{l}
h_{1}=u \sin \theta t_{1}-\frac{1}{2} g t_{1}^{2} \\
t_{1}=\frac{h_{1}+\frac{1}{2} g t_{1}^{2}}{u \sin \theta}
\end{array}$
$h_{2}=u \sin \theta t_{2}-\frac{1}{2} g t_{2}^{2}$
or $t_{2}=\frac{h_{2}+\frac{1}{2} g t_{2}^{2}}{u \sin \theta}$
Divide Eq. (i) by Eq. (ia)
$\begin{array}{l}
\frac{t_{1}}{t_{2}}=\frac{h_{1}+\frac{1}{2} g t_{1}^{2} / u \sin \theta}{h_{2}+\frac{1}{2} g t_{2}^{2} / u \sin \theta} \\
\Rightarrow h_{1} t_{2}-h_{2} t_{1}=\frac{g}{2}\left(t_{1} t_{2}^{2}-t_{1}^{2} t_{2}\right)
\end{array}$
The time of flight of the ball $T=\frac{2 u \sin \theta}{g}=\frac{2}{g}(u \sin \theta) \quad$
$=\frac{2}{g}\left[\frac{h_{1}+\frac{1}{2} g t_{1}^{2}}{t_{1}}\right]=\frac{2}{t_{1}}\left[\frac{h_{1}}{g}+\frac{t_{1}^{2}}{2}\right]$
$=\frac{h_{1}}{t_{1}} \times \frac{2}{g}+t_{1}=\frac{h_{1}}{t_{1}} \times\left(\frac{t_{1} t_{2}^{2}-t_{1}^{2} t_{2}}{h_{1} t_{2}-h_{2} t_{1}}\right)+t_{1}$
$=\frac{h_{1} t_{1} t_{2}^{2}-h_{1} t_{1}^{2} t_{2}+h_{1} t_{1}^{2} t_{2}-h_{2} t_{1}^{3}}{t_{1}\left(h_{1} t_{2}-h_{2} t_{1}\right)}$
$=\left(\frac{h_{1} t_{2}^{2}-h_{2} t_{1}^{2}}{h_{1} t_{2}-h_{2} t_{1}}\right)$
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