#### Answer

The speed of the rock is 19.0 m/s

#### Work Step by Step

We can find the spring constant of the rubber band. Note that the rubber band has a force of 60 N when it is stretched 0.30 meters.
$kx = F$
$k = \frac{F}{x}$
$k = \frac{60~N}{0.30~m}$
$k = 200~N/m$
We can find the energy stored in the rubber band when it is stretched 30 cm.
$U_s = \frac{1}{2}kx^2$
$U_s = \frac{1}{2}(200~N/m)(0.30~m)^2$
$U_s = 9.0~J$
When the rock leaves the rubber band, the kinetic energy will be equal to the energy stored in the rubber band initially. We can find the speed of the rock.
$KE = U_s$
$\frac{1}{2}mv^2 = 9.0~J$
$v^2 = \frac{(2)(9.0~J)}{m}$
$v = \sqrt{\frac{(2)(9.0~J)}{m}}$
$v = \sqrt{\frac{(2)(9.0~J)}{0.050~kg}}$
$v = 19.0~m/s$
The speed of the rock is 19.0 m/s