Sure, here's the solution in LaTeX format:

Given:
Initial velocity $(v_0) = 30 \text{ m/s}$
Angle $(\theta) = 45^\circ$
Acceleration due to gravity on Mars $(g) = 3.71 \text{ m/s}^2$

First, we can find the initial vertical and horizontal components of the velocity:
$v_{0x} = v_0 \cos(\theta) = 30 \cos(45^\circ) \approx 21.21 \text{ m/s}$
$v_{0y} = v_0 \sin(\theta) = 30 \sin(45^\circ) \approx 21.21 \text{ m/s}$

To calculate the maximum height $(H)$ reached by the ball, we can use the following equation:
$v^2 = u^2 + 2as$
where:
$v = \text{final vertical velocity (0 m/s at maximum height)}$
$u = \text{initial vertical velocity} (v_{0y})$
$a = \text{acceleration due to gravity} (-g, \text{because it's acting opposite to the direction of motion})$
$s = \text{vertical displacement} (H)$

$0 = (21.21 \text{ m/s})^2 + 2(-3.71 \text{ m/s}^2)H$

Solving for $H$, we get:
$H \approx (21.21 \text{ m/s})^2 / (2 \cdot 3.71 \text{ m/s}^2) \approx 60.67 \text{ m}$

Next, we can find the time $(t)$ the ball spends in the air using the following equation:
$v = u + at$

$0 = 21.21 \text{ m/s} - 3.71 \text{ m/s}^2 \cdot t$
$t \approx 5.71 \text{ s}$

Now, we can calculate the horizontal distance $(D)$ traveled by the ball using the horizontal component of the initial velocity and the time of flight:
$D = v_{0x} \cdot t = 21.21 \text{ m/s} \cdot 5.71 \text{ s} \approx 121.13 \text{ m}$

So, the maximum height reached by the ball is approximately 60.67 meters, and the final distance from the initial point to the impact point is approximately 121.13 meters.
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