Vertical spring (trampoline) compresion.

1. The problem statement, all variables and given/known data
A student jumping on a trampoline reaches a maximum height of h = 0.96 m. The student has a mass of m = 58 kg.
What is the student’s speed immediately before she reaches the trampoline after the jump in m/s?
answer: 4.338
If, when she lands on the trampoline, she stretches the trampoline down d = 0.75 m, what is the spring constant k in N/m of the trampoline?
h=0.96m
m=58 kg
d= 0.75
v=4.338
2. Relevant equations
KE=(1/2)mv2
F=(1/2)kx
SpringPE=(1/2)kx2
ΔPe=mgh
W=fd cos(Θ)
KE=-PE

3. The attempt at a solution
I have had a gruesome time trying to understand spring related questions.
First I took F=-kx and solved for k, leading to -k=F/x leading to k=mg/x. This is wrong however, because the force of the trampoline is greater than the force exerted by gravity on the person. so in the equation we have 2 unknowns, k and f.
So I took formula KE=-PE
KE=(1/2)mv2
SpringPE=(1/2)kx2
leading to (1/2)mv2=-(1/2)kx2
thus -mv2/x2=k
This didn’t give the correct answer and I’m not sure why. Is KE also supposed to inclued the KE of the falling person?

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