Answer:
v = 16.11 m / s
Explanation:
For this exercise we must use the principle of conservation of energy. We set a reference system on the part of the platform without elongation
starting point. When the spring is compressed
    Em₀ = K_e + U = ½ k x² + m g x ’
final point. The point where it leaves the platform
    Em_f = K = ½ m v²
energy is conserved
     Em₀ = Em_f
     ½ k x² + m g x ’= ½ m v²
     v² = [tex]\frac{k}{m}[/tex]  x² + g x
let's calculate
     v² = [tex]\frac{8700}{55}[/tex]  1.25² + 9.8 1.25
     v² = 247.159 + 12.25 = 259.409
     v = 16.11 m / s