Answer
(a) This is given by the maximum range the ball achieves i.e. $R=120\text{ m}$.
(b) The minimum speed is $v_{min}=24.3\text{ m/s}.$
Work Step by Step
(a) The maximum horizontal distance that the ball can travel is when it is shot at the $45^\circ$ angle above the horizontal when we neglect the air resistance. The total time of the flight is twice the time of the ascending so we have
$$t_f=2\frac{v_0\sin45^\circ}{g}.$$
The maximum range is thus the, constant, horizontal component of the velocity multiplied by this time
$$R=v_xt_f=2\frac{v_0^2\sin45^\circ\cos45^\circ}{g}=\frac{v_0^2}{g}=\frac{(34.4\text{ m/s})^2}{9.81\text{ m/s}^2}=120\text{ m}.$$
(b) Since the horizontal component of the velocity does not change, the minimum speed is achieved when the vertical component is zero which happens when the ball reaches the maximum height. So we have:
$$v_{min}=v_x=v_0\cos45^\circ=34.4\cdot\frac{\sqrt{2}}{2}\text{ m/s}=24.3\text{ m/s}.$$
