How Is the Spring Constant Calculated for an Acrobat's Safe Landing?

[tartio]

Homework Statement

A very short 100 kg acrobat steps off a 10 m high platform and
starts falling. The acrobat falls 9 m, then encounters a 1 m long
spring connected at one end to the ground below. (a) What’s the
spring constant so the acrobat just touches the ground (that is, so the
spring contracts 1 meter)? The spring-with-acrobat system rebounds
and launches the acrobat back into the air. (b) How high off the
ground does the acrobat go above the ground before falling back?

Homework Equations

(1/2)k(x-x0)^2
mgΔh

The Attempt at a Solution

To solve this, wouldn't one set the gravitational potential energy (mgh) to the spring potential energy (1/2kx^2) because the energy converts to one another, then solve for k? 1.96 * 10^4 N/m is the answer I got.

For the second part of the problem, the acrobat will be the same exact height above the ground because the energy is conserved, right?

Am I missing something in this problem? This was unexpectedly simple.

 

[CompuChip]

You are on the right track, but be careful: you wrote Δh in the second formula, not h. What is the difference in this case?

Part two seems to be that simple, unless I'm also missing something :)
Let's hope for the acrobat that there is some loss due to friction (or the spring won't fully extend after having been crushed to length 0) otherwise it may be a long evening for him!