A ball is released from a height h. If t_{1} | vedclass

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(D) For the first half distance $\frac{h}{2}$,the time taken is $t_{1}$:$\frac{h}{2} = \frac{1}{2} g t_{1}^{2} \implies h = g t_{1}^{2}$  (Equation $1

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$t_{2} = (\sqrt{2}-1) t_{1}$

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(D) For the first half distance $\frac{h}{2}$,the time taken is $t_{1}$:$\frac{h}{2} = \frac{1}{2} g t_{1}^{2} \implies h = g t_{1}^{2}$  (Equation $1$)For the total distance $h$,the total time taken is $(t_{1} + t_{2})$:$h = \frac{1}{2} g (t_{1} + t_{2})^{2}$ (Equation $2$)Equating the two expressions for $h$:$g t_{1}^{2} = \frac{1}{2} g (t_{1} + t_{2})^{2}$$2 t_{1}^{2} = (t_{1} + t_{2})^{2}$Taking the square root on both sides:$\sqrt{2} t_{1} = t_{1} + t_{2}$$t_{2} = \sqrt{2} t_{1} - t_{1}$$t_{2} = (\sqrt{2} - 1) t_{1}$

$A$ ball is released from a height $h$. If $t_{1}$ and $t_{2}$ are the times required to complete the first half and the second half of the distance respectively,then choose the correct relation between $t_{1}$ and $t_{2}$.

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