$A$ man stands between two high-rise buildings and blows a whistle. He hears two successive echoes after $0.4 \, s$ and $2.0 \, s$. Calculate the distance between the buildings. (Assume the speed of sound $v = 332 \, m/s$)

(N/A) Let the speed of sound be $v = 332 \, m/s$.
The time taken for the first echo is $t_1 = 0.4 \, s$. The distance of the person from the nearer building $(d_1)$ is given by:
$d_1 = \frac{v \times t_1}{2} = \frac{332 \times 0.4}{2} = 66.4 \, m$.
The time taken for the second echo is $t_2 = 2.0 \, s$. The distance of the person from the farther building $(d_2)$ is given by:
$d_2 = \frac{v \times t_2}{2} = \frac{332 \times 2.0}{2} = 332 \, m$.
The total distance between the two buildings is the sum of the distances from the person to each building:
$D = d_1 + d_2 = 66.4 + 332 = 398.4 \, m$.