To solve this problem, we need to split the initial velocity into its horizontal (Vx) and vertical (Vy) components, and then use the following kinematic equations:

Vertical motion equation for maximum height: Vy^2 = Viy^2 + 2 * a * h
Horizontal motion equation for range: x = Vx * t
Given that the angle is 45 degrees and the initial speed is 30 m/s, we can calculate the horizontal and vertical components of the initial velocity:

Vix = Vi * cos(45°) = 30 * cos(45°) = 30 * 0.7071 ≈ 21.21 m/s
Viy = Vi * sin(45°) = 30 * sin(45°) = 30 * 0.7071 ≈ 21.21 m/s

The acceleration due to gravity on Mars is approximately -3.71 m/s^2. We can now use the vertical motion equation to find the maximum height (h) reached by the ball:

Vy^2 = Viy^2 + 2 * a * h
0 = (21.21 m/s)^2 + 2 * (-3.71 m/s^2) * h

Solving for h, we get:

h = (21.21 m/s)^2 / (2 * 3.71 m/s^2) ≈ 60.67 m

Now, we need to find the time it takes for the ball to reach the ground. We can use the following kinematic equation for vertical motion:

y = Viy * t + 0.5 * a * t^2

Since the ball will reach the ground, y = 0. Plugging in the known values, we have:

0 = 21.21 m/s * t + 0.5 * (-3.71 m/s^2) * t^2

This is a quadratic equation in the form of at^2 + bt + c = 0, with a = -0.5 * 3.71 m/s^2, b = 21.21 m/s, and c = 0. To find the time, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

Since we're only interested in the positive value for time, we choose the positive sign:

t = (-21.21 + √(21.21^2)) / (-2 * 0.5 * 3.71) ≈ 11.47 s

Now, we can use the horizontal motion equation to find the final distance (x) from the initial point to the impact point:

x = Vx * t = 21.21 m/s * 11.47 s ≈ 243.17 m

Thus, the maximum height reached by the ball is approximately 60.67 meters, and the final distance from the initial point to the impact point with Mars' surface is approximately 243.17 meters.
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