Answer
The trampoline stretches a distance of $8.65~cm$
Work Step by Step
By conservation of energy, the gravitational potential energy at maximum height above the starting point will be equal to the elastic potential energy stored in the trampoline initially. We can find the effective spring constant of the trampoline:
$\frac{1}{2}kx^2 = mg(h+x)$
$k = \frac{2mg(h+x)}{x^2}$
$k = \frac{(2)(52~kg)(9.80~m/s^2)(2.5~m+0.75~m)}{(0.75~m)^2}$
$k = 5889~N/m$
We can find the distance the trampoline stretches when the gymnast stands at rest:
$kx = mg$
$x = \frac{mg}{k}$
$x = \frac{(52~kg)(9.80~m/s^2)}{5889~N/m}$
$x = 0.0865~m = 8.65~cm$
The trampoline stretches a distance of $8.65~cm$.
