It is given that $v=40 \mathrm{m} {\mathrm{s}}^{-1}$.

The components of the velocity will be $$\begin{gathered} v_y=v\sin 30°=\frac{40 \mathrm{m} {\mathrm{s}}^{-1}}{2}=20 \mathrm{m} {\mathrm{s}}^{-1} \\ {v}_x\mathit{=}v\cos 30°=\frac{40\sqrt{3}}{2} \mathrm{m} {\mathrm{s}}^{-1}=20\sqrt{3} \mathrm{m} {\mathrm{s}}^{-1} \end{gathered}$$

The time taken to reach maximum height is $T=\frac{u\sin 30°}{g}=\frac{20 \mathrm{m} {\mathrm{s}}^{-1}}{10 \mathrm{m} {\mathrm{s}}^{-2}}=2 \mathrm{s}$

Hence, at $T=2 \mathrm{s},\mathit{ }\mathit{\text{as}}\mathit{ }{v}_y=0 \mathrm{m} {\mathrm{s}}^{-1} \text{therefore} v_{net}=v_x=20\sqrt{3} \mathrm{m} {\mathrm{s}}^{-1}$.