Planet X has three times the free-fall acceleration of Earth.
a. A ball is thrown vertically upward with the same initial velocity on Earth and on X. How does the maximum height reached by the ball on X compare to the maximum height on Earth? Justify your answer with both words and kinematics calculations. Your answer should state "The ball goes times higher on X."
b. Now, a ball is thrown vertically upward on X with three times the initial velocity of an identical ball on earth. How does the maximum height reached by the ball on X compare to the maximum height on Earth? Justify your answer with both words and kinematics calculations. Your answer should state "The ball goes __ times higher on X."

As the initial velocity is the same than on Earth, but the free-fall acceleration is three times larger, this means that the only net force acting on the ball (gravity) will be three times larger, so it is clear that the ball will reach to a lower height, as it will slowed down more quickly.
Kinematically, as we know that the speed becomes zero when the ball reaches to the maximum height, we can use the following kinematic equation:

[tex]v_{f} ^{2} - v_{o}^{2} = 2* \Delta h* g[/tex]

since vf = 0, solving for Δh, we have:

[tex]\Delta h = h_{max} =\frac{v_{o} ^{2}}{2*g} (1)[/tex]

if v₀ₓ = v₀E, and gₓ = 3*gE, replacing in (1), we get:

Δhₓ = 1/3 * ΔhE

which confirms our intuitive reasoning.

b)

Now, if the initial velocity is three times larger than the one on Earth, even the acceleration due to gravity is three times larger, we conclude that the ball will go higher than on Earth.
We can use the same kinematic equation as in (1) replacing Vox by 3*VoE, as follows:

[tex]\Delta h = h_{max} =\frac{(3*v_{o}) ^{2}}{2*3*g} (2)[/tex]

Replacing the right side of (1) in (2), we get:

Δhx = 3* ΔhE

which confirms our intuitive reasoning also.