#### Answer

The plane's landing speed was 60 m/s

#### Work Step by Step

The plane's kinetic energy when it lands will be equal to the energy $U_s$ stored in the spring. We can find the landing speed of the plane. Therefore;
$KE = U_s$
$\frac{1}{2}mv^2=\frac{1}{2}kx^2$
$v^2=\frac{kx^2}{m}$
$v=\sqrt{\frac{kx^2}{m}}$
$v=\sqrt{\frac{(60,000~N/m)(30~m)^2}{15,000~kg}}$
$v = 60~m/s$
The plane's landing speed was 60 m/s.