The motion of the ball on the vertical axis is an accelerated motion, with acceleration
[tex]a=g=-9.81 m/s^2[/tex]
The following relationship holds for an uniformly accelerated motion:
[tex]2aS=v_f^2 - v_i^2[/tex]
where S is the distance covered, vf the final velocity and vi the initial velocity.
If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
[tex]v_f =0[/tex]
So we can rewrite the equation as
[tex]2(-9.81 m/s^2) h=-v_i^2[/tex]
from which we can isolate h
[tex]h= \frac{v_i^2}{19.62} [/tex] (1)
Now let's assume that [tex]v_i[/tex] is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: [tex]2v_i[/tex]. So the maximum height of the second ball is
[tex]h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62} [/tex] (2)
Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
