physics help please!!?

Question

an 80kg man jumps from a height of 2.50meters onto a platform down a mounted on springs. As the springs compress,. he pushes the platform down a maximum distance of .240m below its initial position, and then it rebounds. the platform and springs have neglible mass.
a) what is the man's speed at the instant he depresses the platform .120m?
b) if the man steps gently onto the platform, what maximum distance would he push it down?

Answer 1

Pe =mgh
Pe= 80 x 9.81 x 2.5= 1962 Joules


b)
Ps=0.5kx^2
Ke=0.5mv^2

Ke(at x= .120m) + Ps(at x=.120m)=Ps(max)
.5mv^2 + .5 k (.120)^2 = 5 k (.240)^2

v= sqrt(k[(240)^2 - .120)^2]/m)
k- spring const?
Ps=Ke=Pe (all max values)

.5kx^2=Pe(max)
k=2Pe(max)/(x^2)= 2 x 1962 /(.24)^2=
k=68125N/m^2

v= sqrt(68125[(0.240)^2 - 0.120)^2]/80)
v=6.1m/s

b)F=-kx
x=-F/k=-mg/k
x=-80 x 9.81 /68125=0.01152 m
or x=1.15 cm

Answer 2

a) His kenetic energy the instant he hits will be equal to the gravitational potential at the begining so:
Ke = Pe = 1/2mv² = mgh

When its depressed .120 or half way, he will have half of this kenetic energy, energy is conserved in compressing the spring. Equate this energy with kenetic energy ( 1/2mv² ) to solve for v.

1/2mgh = 1/2mv²
gh = v²
v = sqrt(gh)
v ~= 4.95 m/s

b) Use Hooks Law:
F = kx

Solve for k again using conservation of energy, the equation for gravitational potential energy, and elastic potential energy.

mgh = 1/2kx²
(2mgh)/x² = k
k ~=68055.55
That doesn't seem to right but whatever I'll go with it

Back to hooks law
F = kx

The only force would be gravity on the man (m*g) and the k is (2mgh/x²) so

mg = kx
x = mg/k
x = 0.01152m or 1.152cm