Acceleration $\mathrm{g}$ is vertically downward
To make velocity perpendicular to acceleration, velocity must be horizontal. This is possible at maximum height.

For this
${\mathrm{v}}_{\mathrm{y}}=0$
${\mathrm{u}}_{\mathrm{y}}=\mathrm{u}\sin \mathrm{\theta }$
${\mathrm{u}}_{\mathrm{y}}=20\sin 30°=10\mathrm{ }\mathrm{m} {\mathrm{s}}^{-2}$
$\mathrm{g}=-10\mathrm{ }{\mathrm{ms}}^{-2}$

We know that
${\mathrm{v}}_{\mathrm{y}}={\mathrm{u}}_{\mathrm{y}}+\mathrm{g}\mathrm{t}$
By putting all value, we get $\mathrm{t}=1\mathrm{ }\text{s}$