Answer
(a) The cliff is $260~m$ away.
(b) The intensity of the music arriving at the cliff is $1.5\times 10^{-10}~W/m^2$
Work Step by Step
(a) Let $r_c$ be the distance to the cliff. The sound takes $1.5~s$ to travel a distance of $2~r_c$. We can find $r_c$:
$2~r_c = v~t$
$r_c = \frac{v~t}{2}$
$r_c = \frac{(343~m/s)(1.5~s)}{2}$
$r_c = 257.25~m$
$r_c = 260~m$
The cliff is $260~m$ away.
(b) We can write an expression for the intensity 1.0-m away:
$I_1 = \frac{P}{A}$
$I_1 = \frac{P}{4\pi~r_1^2}$
We can write an expression for the intensity at the cliff:
$I_c = \frac{P}{A}$
$I_c = \frac{P}{4\pi~r_c^2}$
We can divide the two equations to find $I_c$:
$\frac{I_c}{I_1} = \frac{\frac{P}{4\pi~r_c^2}}{\frac{P}{4\pi~r_1^2}}$
$I_c = \frac{r_1^2}{r_c^2}~\times I_1$
$I_c = \frac{(1.0~m)^2}{(257.25~m)^2}\times (1.0\times 10^{-5}~W/m^2)$
$I_c = 1.5\times 10^{-10}~W/m^2$
The intensity of the music arriving at the cliff is $1.5\times 10^{-10}~W/m^2$
