To calculate the maximum height and the final distance of a ball thrown with an initial velocity and angle, we can use the kinematic equations and basic trigonometry. However, it is important to note that the acceleration due to gravity on Mars is different from that on Earth. On Mars, the gravitational acceleration is approximately 3.71 m/s².

Given:
Initial velocity (v₀) = 30 m/s
Angle (θ) = 45 degrees
Acceleration due to gravity on Mars (g) = 3.71 m/s²

First, we can find the initial vertical and horizontal components of the velocity:
v₀x = v₀ * cos(θ) = 30 * cos(45) = 30 * 0.7071 ≈ 21.21 m/s
v₀y = v₀ * sin(θ) = 30 * sin(45) = 30 * 0.7071 ≈ 21.21 m/s

To calculate the maximum height (H) reached by the ball, we can use the following equation:
v² = u² + 2as
where:
v = final vertical velocity (0 m/s at maximum height)
u = initial vertical velocity (v₀y)
a = acceleration due to gravity (-g, because it's acting opposite to the direction of motion)
s = vertical displacement (H)

0 = (21.21 m/s)² + 2*(-3.71 m/s²)*H

Solving for H, we get:
H ≈ (21.21 m/s)² / (2 * 3.71 m/s²) ≈ 60.67 m

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

0 = 21.21 m/s - 3.71 m/s² * t
t ≈ 5.71 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 = v₀x * t = 21.21 m/s * 5.71 s ≈ 121.13 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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