$A$ cricket ball is thrown at a speed of $28 \; m/s$ in a direction $30^{\circ}$ above the horizontal. Calculate:
$(a)$ the maximum height,
$(b)$ the time taken by the ball to return to the same level,and
$(c)$ the distance from the thrower to the point where the ball returns to the same level.

(N/A) The maximum height is given by $h_{m} = \frac{(v_{0} \sin \theta_{0})^{2}}{2g} = \frac{(28 \sin 30^{\circ})^{2}}{2 \times 9.8} = \frac{14 \times 14}{19.6} = 10.0 \; m$.
$(b)$ The time taken to return to the same level (time of flight) is $T_{f} = \frac{2 v_{0} \sin \theta_{0}}{g} = \frac{2 \times 28 \times \sin 30^{\circ}}{9.8} = \frac{28}{9.8} \approx 2.86 \; s$ (rounded to $2.9 \; s$).
$(c)$ The horizontal range is $R = \frac{v_{0}^{2} \sin 2\theta_{0}}{g} = \frac{28^{2} \times \sin 60^{\circ}}{9.8} = \frac{784 \times 0.866}{9.8} \approx 69.28 \; m$ (rounded to $69 \; m$).