Answer
$$3.30\times10^{4}\;m$$
Work Step by Step
The Mach cone angle is
$\sin\theta=\frac{v}{v_s}$
or, $\sin\theta=\frac{v}{1.25v}=0.8$
or, $\theta=53.13^{\circ}$
The sonic boom reaches the man on the ground $1.00\;min$ after the plane passes directly overhead.
The distance covered by the plane in $1.00\;min$ is
$x=v_st$
or,$ x=1.25vt$
or, $x=1.25\times330\times60\;m$
or, $x=24750 \;m$
Let $H$ be the altitude of the plane.
$\therefore\;\;H=x\tan\theta$
or, $H=24750\tan53.13^{\circ}\;m$
or, $H=3.30\times10^{4}\;m$
Therefore, the altitude of the plane is $3.30\times10^{4}\;m$
