Question

Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the first ball. how does the maximum height of the two balls compare? two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the first ball. how does the maximum height of the two balls compare? the maximum height of the second ball is two times that of the first ball. the maximum heights of the two balls are equal. the maximum height of the second ball is four times that of the first ball. the maximum height of the second ball is 1.41 times that of the first ball. the maximum height of the second ball is eight times that of the first ball.

Answer 1

The motion of the ball on the vertical axis is an accelerated motion, with acceleration 
$a=g=-9.81 m/s^2$
The following relationship holds for an uniformly accelerated motion:
$2aS=v_f^2 - v_i^2$
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:
$v_f =0$
So we can rewrite the equation as
$2(-9.81 m/s^2) h=-v_i^2$
from which we can isolate h
$h= \frac{v_i^2}{19.62}$ (1)

Now let's assume that $v_i$ is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: $2v_i$. So the maximum height of the second ball is
$h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62}$ (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.