A ball is thrown horizontally from the top of a vertical tower and strikes the ground 50m from the bottom. If the tower is 45m high, calculate the time taken to hit the ground.

Initial vertical velocity, $u_y = 0 \text{ m/s}$ (since it's thrown horizontally).  
Vertical displacement, $s_y = h = 45 \text{ m}$.  
Use the kinematic equation: $s_y = u_y t + \frac{1}{2}gt^2$.
$$
45 \text{ m} = (0 \text{ m/s})t + \frac{1}{2}(9.8 \text{ m/s}^2)t^2
$$
$$
45 = 4.9t^2
$$

$$
t^2 = \frac{45}{4.9}
$$
$$
t^2 \approx 9.18367
$$
$$
t = \sqrt{9.18367}
$$
$$
t \approx 3.03 \text{ s}
$$
The time taken by the ball to hit the ground is $\boxed{\text{3.03 s}}$.

Horizontal displacement, $s_x = 50 \text{ m}$.  
Time of flight, $t \approx 3.03 \text{ s}$ (from part i).  
Horizontal velocity, $u_x$, is constant.  
Use the equation: $s_x = u_x t$.
$$
50 \text{ m} = u_x (3.03 \text{ s})
$$

$$
u_x = \frac{50 \text{ m}}{3.03 \text{ s}}
$$
$$
u_x \approx 16.5 \text{ m/s}
$$
The initial horizontal velocity of the ball is $\boxed{\text{16.5 m/s}}$.

Initial vertical velocity, $u_y = 0 \text{ m/s}$.  
Time of flight, $t \approx 3.03 \text{ s}$.  
Acceleration, $g = 9.8 \text{ m/s}^2$.  
Use the kinematic equation: $v_y = u_y + gt$.
$$
v_y = 0 \text{ m/s} + (9.8 \text{ m/s}^2)(3.03 \text{ s})
$$

$$
v_y \approx 29.694 \text{ m/s}
$$
$$
v_y \approx 29.7 \text{ m/s}
$$
The vertical velocity of the ball just before striking the ground is $\boxed{\text{29.7 m/s}}$.