Answer
(a) $\theta = 28.4^{\circ}$
(b) A person on the ground will hear the shock wave 23.2 seconds after the plane passes overhead.
Work Step by Step
(a) We can find the angle that the shock wave makes.
$sin(\theta) = \frac{v_{snd}}{v_{obj}}$
$\theta = arcsin(\frac{v_{snd}}{v_{obj}})$
$\theta = arcsin(\frac{v_{snd}}{2.1v_{snd}})$
$\theta = arcsin(\frac{1}{2.1})$
$\theta = 28.4^{\circ}$
(b) We can find the distance $d_s$ traveled by the shock wave.
$sin(\theta) = \frac{h}{d_s}$
$d_s = \frac{h}{sin(\theta)}$
$d_s = \frac{6500~m}{sin(28.4^{\circ})}$
$d_s = 13,670~m$
We can find the time $t_s$ for the shock wave to travel this distance.
$t_s = \frac{13,670~m}{343~m/s} = 39.9~s$
We can find the distance $d_p$ traveled by the plane.
$tan(\theta) = \frac{h}{d_p}$
$d_p = \frac{h}{tan(\theta)}$
$d_p = \frac{6500~m}{tan(28.4^{\circ})}$
$d_p = 12,020~m$
We can find the time $t_p$ for the plane to travel this distance.
$t_p = \frac{12,020~m}{(343~m/s)(2.1)} = 16.7~s$
The time difference is $39.9~s-16.7~s$ which is $23.2~s$. Therefore, a person on the ground will hear the shock wave 23.2 seconds after the plane passes overhead.
