A projectile is fired with a velocity of $20$ m s$^{-1}$ at an angle of $30^\circ$ to the horizontal. Find its time of flight. (Take $g = 10$ m s$^{-2}$.)

Time of flight $T = \dfrac{2u\sin\theta}{g} = \dfrac{2 \times 20 \times \sin 30^\circ}{10} = \dfrac{2 \times 20 \times 0.5}{10} = 2$ s. Using $\cos 30^\circ$ gives $2\sqrt{3}$, and omitting the factor 2 gives the time to maximum height, 1 s.