A projectile is projected at 30 ° from horizontal with initial...

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

The components of the velocity will be ${v}_{y} = vsin30 ° = \dfrac{40m{s}^{- 1}}{2} = 20m{s}^{- 1}{v}_{x} = vcos30 ° = \dfrac{40\sqrt{3}}{2}m{s}^{- 1} = 20\sqrt{3}m{s}^{- 1}$

The time taken to reach maximum height is $T = \dfrac{usin30 °}{g} = \dfrac{20m{s}^{- 1}}{10m{s}^{- 2}} = 2s$

Hence, at $T = 2s, \text{as}{v}_{y} = 0m{s}^{- 1}\text{therefore}{v}_{net} = {v}_{x} = 20\sqrt{3}m{s}^{- 1}$.