Problem: A clown of mass M = 76.8 kg jumps off a tower at height H = 17.9 m above a net that is stretched horizontally. The net acts like a spring with spring constant k = 16900 N/m. How far will the net stretch before the clown comes instantaneously to rest?
U = mgh
KE = (1/2)mv^2
Energy (Total) of a Spring (Under Simple Harmonic Motion) = (1/2)kA^2, where A = amplitude in meters.
The Attempt at a Solution
Initially: the clown possesses U = mgh Joules.
After: Right before falling down on the net, he possesses U = KE = (1/2)mv^2 Joules, by Conservation of Energy. (We will call this point (arbitrary), U = O.
At the point where the net stretches:
E = 1/2(k)A^2
Since energy is conserved (none lost to air friction, etc.), mgh = (1/2)kA^2. Solving for A, I get 1.2 meters. (which was marked incorrect)
If there is any hint/tips, I would greatly appreciate it!