A cricket ball is thrown at a speed of $28\; m /s$ in a direction $30^o$ 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

$(a)$ The maximum height is given by

$h_{m} =\frac{\left(v_{o} \sin \theta_{ o }\right)^{2}}{2 g}$$=\frac{\left(28 \sin 30^{\circ}\right)^{2}}{2(9.8)} m$$=\frac{14 \times 14}{2 \times 9.8}=10.0 m$

$(b)$ The time taken to return to the same level is

$T_{f}=\left(2 v_{ o } \sin \theta_{ o }\right) / g$$=\left(2 \times 28 \times \sin 30^{\circ}\right) / 9.8$

$=28 / 9.8 s =2.9 s$

$(c)$ The distance from the thrower to the point where the ball returns to the same level is

$R=\frac{\left(v_{ o }^{2} \sin 2 \theta_{ o }\right)}{g}$$=\frac{28 \times 28 \times \sin 60^{\circ}}{9.8}=69 m$