We can solve this problem using basic kinematic equations. First, we'll find the initial velocity components in the vertical (Vy) and horizontal (Vx) directions. Given the angle (45 degrees) and the initial speed (30 m/s), we can use the following equations:

Vy = V * sin(θ)
Vx = V * cos(θ)

where V is the initial speed, θ is the angle, and Vy and Vx are the vertical and horizontal components, respectively.

For a 45-degree angle, sin(45) = cos(45) = √2/2.

Vy = 30 * (√2/2) = 15√2 m/s
Vx = 30 * (√2/2) = 15√2 m/s

Since Mars has a different gravitational acceleration (g) than Earth, we'll use g_Mars = 3.721 m/s². To calculate the maximum height (H) reached by the ball, we'll use the following equation:

Vf² = Vi² - 2 * g * H

where Vf is the final vertical velocity (0 m/s at the maximum height), Vi is the initial vertical velocity (Vy), g is the gravitational acceleration on Mars, and H is the maximum height.

0 = (15√2)² - 2 * 3.721 * H

Solving for H, we get:

H = (15√2)² / (2 * 3.721) = 180 / 3.721 ≈ 48.34 m

Now, we'll find the time (t) it takes for the ball to reach its maximum height using the following equation:

Vf = Vi - g * t

0 = 15√2 - 3.721 * t

t = (15√2) / 3.721 ≈ 5.70 s

Since the time to reach the maximum height is half the total time in the air, the total time (T) is:

T = 2 * t ≈ 2 * 5.70 = 11.4 s

Now we can find the final distance (D) from the initial point to the impact point with the Mars surface using the horizontal velocity component (Vx) and the total time (T):

D = Vx * T = 15√2 * 11.4 ≈ 242.54 m

So, the maximum height reached by the ball is approximately 48.34 m, and the final distance from the initial point to the impact point with the Mars surface is approximately 242.54 m.
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