Answer:
The maximum length during the motion is 
Explanation:
From the question we are told that
The mass is 
The vertical spring length is 
The unstretched length is 
The initial speed is 
The new length of the spring 
The spring constant k is mathematically represented as

Where F is the force applied 
y is the difference in weight which is 
The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)
Now substituting values accordingly


The elastic potential energy is given as 
where D is this the is the displacement
Since Energy is conserved the total elastic potential energy would be

Substituting value accordingly




So to obtain total length we would add the unstretched length
So we have
