A rock of mass m is thrown straight up into the air with initial speed |v0 | and initial position y = 0 and it rises up to a maximum height of y = h. A second rock with mass 2m (twice the mass of the original) is thrown straight up with an initial speed of 2|v0 |. What maximum height does the second rock reach?

Case 1:

mass = m

initial velocity = vo

final velocity = 0

height = y

Use third equation of motion

v² = u² - 2as

0 = vo² - 2 g y

y = vo² / 2g ... (1)

Case 2:

mass = 2m

initial velocity = 2vo

final velocity = 0

height = y '

Use third equation of motion

v² = u² - 2as

0 = 4vo² - 2 g y'

y ' = 4vo² / 2g

y' = 4 y

Thus, the second rock reaches the 4 times the distance traveled by the first rock.