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The acceleration due to gravity (constant) can be measured by projecting a body upward and measuring the time that it takes to pass two given lines in both directions (upward motion and downward motion). If the time the body takes to pass a horizontal line A in both directions (from $A_{1} t o A_{2}$ ) is $T_{A}$ , and the time to go by a second line B in both directions is $T_{B}$ ?(from $B_{1}$ , to $B_{2}$ ). Then the acceleration due to gravity is

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$\frac{8 h}{T_{A}^{2} - T_{B}^{2}}$

$\frac{8 h}{T_{A}^{2} + T_{B}^{2}}$

$\frac{8 h}{T_{A}^{2} T_{B}^{2}}$

$\frac{8 h T_{A} T_{B}}{T_{A}^{2} T_{B}^{2}}$

answer is A.

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Detailed Solution

y-height of line ‘A’ is ‘y’

From the ground

$y = u t - \frac{1}{2} g t^{2}$

$\frac{g t^{2}}{2} - u t + y = 0$ ------(1)

$t = \frac{u \pm \sqrt{u^{2} - 4 (\frac{g}{2}) y}}{2 (\frac{g}{2})}$ ------(2)

Let $t_{1}$ and $t_{2}$ are time the body passes the two points of the same horizontal line (from the start)

Then $T_{A} = t_{2} - t_{1}$

(2) $\Rightarrow$ $t_{1} = \frac{u - \sqrt{u^{2} - 2 g y}}{g}, t_{2} = \frac{u + \sqrt{u^{2} - 2 g y}}{g}$ \

$T_{A}^{2} = {(t_{2} - t_{1})}^{2}$

$T_{A}^{2} = {(\frac{u + \sqrt{u^{2} - 2 g y}}{g} - \frac{(u - \sqrt{u^{2} - 2 g y})}{g})}^{2}$

= ${(\frac{2}{g} \sqrt{u^{2} - 2 g y})}^{2}$

= $T_{A}^{2} = \frac{4}{g^{2}} (u^{2} - 2 g y)$

$= similarly, T_{B}^{2} = \frac{4}{g^{2}} (u^{2} - 2 g (h + y))$

= $T_{A}^{2} - T_{B}^{2} = \frac{4}{g^{2}} [u^{2} - 2 g y - u^{2} + 2 g h + 2 g y]$

= $T_{A}^{2} - T_{B}^{2} = \frac{8 h}{g}$

$g = \frac{8 h}{T_{A}^{2} - T_{B}^{2}}$

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