A body is thrown vertically upward with a velocity of 40ms ^-1 . The velocity at the half of the maximum height would be

Initial velocity  $=40m{s}^{-1}$
Maximum height $=\frac{{\mathrm{u}}^2}{2\mathrm{g}}=\frac{40\times 40}{2\times 10}=80\mathrm{m}$
Half of the maximum height $=\frac{80}{2}=40\mathrm{m}$
$\begin{array}{l}{\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as}\\ {\mathrm{v}}^2-(40{)}^2=2(-10\left)\right(40)\\ {\mathrm{v}}^2=-800+1600\\ {\mathrm{v}}^2=800\\ \begin{array}{l}\mathrm{v}=\sqrt{800}\\ \mathrm{v}=20\sqrt{2}{\mathrm{ms}}^{-1}\end{array}\end{array}$