A particle is projected from the ground. If the equation of the trajectory is $\mathrm{y}=\mathrm{x}-{\mathrm{x}}^2/2$, then the time of flight is

$\mathrm{y}=\mathrm{x}-\frac{1}{2}{\mathrm{x}}^2\Rightarrow \mathrm{y}=0$ when $\mathrm{x}=2$. Thus, the path of the projectile is as shown in figure. Clearly, the height is maximum when $\mathrm{x}=1$. The maximum height H is given by

$\mathrm{H}=1-\frac{1}{2}{\left(1\right)}^2=0.5$ Time taken to travel from the highest point to the ground is

$\mathrm{t}=\sqrt{\frac{2\mathrm{H}}{\mathrm{g}}}=\sqrt{\frac{2\times \left(1/2\right)}{\mathrm{g}}}=\frac{1}{\sqrt{\mathrm{g}}}\Rightarrow \mathrm{Time} \mathrm{of} \mathrm{flight}=2\mathrm{t}=2/\sqrt{\mathrm{g}}$