Answer
The plane is a horizontal distance of 16 km away, at a height of 9.5 km.
Work Step by Step
See the figure provided in the textbook for this problem.
Find the angle of the shock wave, using equation 12-5.
$$\theta = sin^{-1}\frac{v_{sound}}{v_{obj}}= sin^{-1}\frac{v_{sound}}{2.0v_{sound}}= sin^{-1}\frac{1}{2.0}=30^{\circ}$$
Now find the horizontal distance that the plane has traveled when the shock wave reaches the observer.
$$tan\theta=\frac{9500m}{L}$$
$$L=\frac{9500m}{tan30^{\circ}}\approx 16\;km$$