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The figure shows four points A, B,C,D on the path of a projectile. The lines AD and BC are horizontal and the vertical distance between these lines is h.
If $T_{1}$ is the time taken by the projectile in moving from A to D and $T_{2}$ is the time taken by the projectile in moving from B to C, then the acceleration g due to Earth's gravity equals

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$g = \frac{8 h}{T_{1}^{2} + T_{2}^{2}}$

$g = \frac{4 h}{{(T_{1} - T_{2})}^{2}}$

$g = \frac{8 h}{T_{1}^{2} - T_{2}^{2}}$

$g = \frac{4 h}{{(T_{1} + T_{2})}^{2}}$

answer is C.

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Detailed Solution

Let v and v' be the vertical velocity components at A and B, respectively. Then

$T_{1} = \frac{2 v}{g} and T_{2} = \frac{2 v^{'}}{g}$

Also $v^{2} - v'^{2} = 2 g h$ . Hence, we have

$\begin{array}{ll} \frac{1}{4} g^{2} (T_{1}^{2} - T_{2}^{2}) = 2 g h \\ \Rightarrow g = \frac{8 h}{T_{1}^{2} - T_{2}^{2}} \end{array}$

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