A 2.2 kg bowling ball is dropped vertically from a height of 1.5 m onto a long compression spring (k = 150 N/m) placed vertically underneath it. What is the maximum compression of the spring?

Answer 1

loss of potential energy by ball = m g (1.5 + x)

gain of potential energy by spring = (1/2)(150)x^2
so
2.2 * 9.81 *(1.5+x) = 75 x^2

21.6 x + 32.4 = 75 x^2
or
75 x^2 - 21.6 x - 32.4 = 0
x = [21.6 +/-sqrt(467+9720) ]/150
x = [ 21.6 + 101 ] /150
x = .817 meter

Answer 2

Damon's post is correct.