A cricket ball thrown across a field is at he | vedclass

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(A) The vertical displacement $y$ of a projectile at time $t$ is given by $y = (u \sin 	heta)t - \frac{1}{2}gt^2$.For heights $h_1$ and $h_2$ at time

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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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(A) The vertical displacement $y$ of a projectile at time $t$ is given by $y = (u \sin 	heta)t - \frac{1}{2}gt^2$.For heights $h_1$ and $h_2$ at times $t_1$ and $t_2$:$h_1 = (u \sin 	heta)t_1 - \frac{1}{2}gt_1^2 \implies \frac{h_1}{t_1} = u \sin 	heta - \frac{1}{2}gt_1 \quad (1)$$h_2 = (u \sin 	heta)t_2 - \frac{1}{2}gt_2^2 \implies \frac{h_2}{t_2} = u \sin 	heta - \frac{1}{2}gt_2 \quad (2)$Subtracting $(2)$ from $(1)$:$\frac{h_1}{t_1} - \frac{h_2}{t_2} = \frac{1}{2}g(t_2 - t_1) \implies \frac{h_1 t_2 - h_2 t_1}{t_1 t_2} = \frac{1}{2}g(t_2 - t_1)$$\frac{g}{2} = \frac{h_1 t_2 - h_2 t_1}{t_1 t_2 (t_2 - t_1)} \quad (3)$The time of flight $T$ is given by $T = \frac{2u \sin 	heta}{g}$. From $(1)$, $u \sin 	heta = \frac{h_1}{t_1} + \frac{1}{2}gt_1$.$T = \frac{2}{g} \left( \frac{h_1}{t_1} + \frac{1}{2}gt_1 ight) = \frac{2h_1}{gt_1} + t_1$.Substituting $\frac{g}{2}$ from $(3)$ into the expression for $T$:$T = \frac{h_1}{t_1} \left( \frac{t_1 t_2 (t_2 - t_1)}{h_1 t_2 - h_2 t_1} ight) + t_1 = \frac{h_1 t_2 (t_2 - t_1) + t_1 (h_1 t_2 - h_2 t_1)}{h_1 t_2 - h_2 t_1}$$T = \frac{h_1 t_2^2 - h_1 t_1 t_2 + h_1 t_1 t_2 - h_2 t_1^2}{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}$.

$A$ 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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