Answer
$k=2373N/m$
Work Step by Step
1) The horizontal speed of the performer is $$v_x=\frac{26.8m}{2.14s}=12.52m/s$$
Because the performer is fired at $\theta=40^o$, the performer's speed as he leaves the cannon is $$v=\frac{12.52m/s}{\cos40^o}=16.35m/s$$
2) At start, the bands are stretched by $x_0=3m$ and the performer's speed $v_0=0$ and height $h_0=0$.
Then, as the bands are released, $x_f=0$, the performer acquires a speed of $v_f=16.35m/s$. Also, we assume the bands' stretching length is the cannon's length, so the height of the performer as he leaves the cannon is $h_f=(3m)\sin40=1.93m$. Therefore, $$\frac{1}{2}kx_0^2=\frac{1}{2}mv_f^2+mgh_f$$ $$k=\frac{mv_f^2+2mgh_f}{x_0^2}$$
The performer's mass $m=70kg$, so $$k=2373N/m$$
