A stone falls freely from rest a height $h$ and it travels a distance of$\dfrac{h}{2}$ in the last second.The time of journey is

Let, $t=$ time taken $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$

$\begin{array}{l} h=\frac{1}{2} g t^{2} \\ \frac{h}{2}=\frac{1}{2} g(t-1)^{2} \\ h=g \times\left(t^{2}-2 t+1\right) \\ \frac{1}{2} g t^{2}=g\left(t^{2}-2 t+1\right) \\ \frac{1}{2} t^{2}=t^{2}-2 t+1 \\ \Rightarrow \frac{1}{2} t^{2}-2 t+1=0 \\ \Rightarrow t^{2}-4 t+2=0 \\ t=\frac{4 \pm \sqrt{16-8}}{2 \times 1} \\ \begin{aligned} = & 2+\sqrt{2}, 2-\sqrt{2} \\ 2-\sqrt{2} & \text { is invalid }[t>2-\sqrt{2}] \end{aligned} \\ \therefore \text { total time } t=2+\sqrt{2} S \text { (Ans) } \end{array}$