I will show my work solving this problem because it helps me compute it. The final answer will be at the bottom in bold.

First we break the initial velocity of the ball into its horizontal and vertical components:

v0x = v0 * cos(45°) = 30 * sqrt(2)/2 = 21.2 m/s (horizontal component)
v0y = v0 * sin(45°) = 30 * sqrt(2)/2 = 21.2 m/s (vertical component)

On Mars, the acceleration due to gravity is approximately 3.71 m/s^2. At the maximum height, the vertical component of the velocity is zero. Therefore, we calculate the time it takes for the ball to reach its maximum height using the equation:

0 = v0y + a*t = 21.2 + -3.71 * t

t = 21.2 / 3.71 = 5.71

We calculate maximum height reached by the ball using the equation:

y = a * t^2 / 2 + v0y * t = -3.71 * 5.71^2 / 2 + 21.2 * 5.71 = -60.5 + 121 = 60.5

Now we need to find the time it takes for the ball to hit the surface of Mars. Once again we use the equation:

y = a * t^2 / 2 + v0y * t

But this time y is 0 and we must solve for t, so we have

0 = a * t^2 / 2 + v0y * t = -3.71 * t^2 / 2 + 21.2 * t = -1.855 * t^2 + 21.2 * t

t = 21.2 / 1.855 = 11.43

Finally, we can find the distance traveled by the ball using the equation:

x = v0x * t = 21.2 * 11.43 = 242

To summarize:

- Therefore, the maximum height reached by the ball is **60.5 m**
- The final distance from the initial point to the impact point with Mars surface is **242 m**

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Keep in mind that I am an LLM, not a calculator, so my results are likely to have errors. I have showed all my work above, so I suggest you double-check and correct any computation mistakes. Hopefully this provides you with enough info to solve the problem, but if you need more help, don't hesitate to ask me clarification questions
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