A person standing at a distance x in front of a cliff fires a gun. Another person B standing behind the person A at a distance y from the cliff hears two sounds of the fired shot after 2 s and 3 s respectively. Calculate x and y (take speed of sound 320 ms^-1)

Given,

speed of sound 320 ms^-1

As we know,

$\text {Speed} = \dfrac{\text {distance}}{\text{time}}$

The person B hears two sounds of the fired shot, the first one is direct from the gun while another sound comes after reflection from the cliff

In the case of first sound for person B, when t = 2s (i.e., when the gun was fired)

Substituting the values in the formula above, we get,

$320 = \dfrac{\text {y - x}}{\text{2}} \\[0.5em] \text {y - x} = 640 \\[0.5em]$

Hence, y - x = 640 m     [Equation 1]

In the case of second sound for person B, when t = 3s (i.e. after reflection from the cliff)

Substituting the values in the formula above, we get,

$320 = \dfrac{\text {y + x}}{\text{3}} \\[0.5em] \text {y + x} = 960 \\[0.5em]$

Hence, y + x = 960 m     [Equation 2]

Adding equation (1) and (2), we get,

y - x = 640     [Equation 1]

y + x = 960     [Equation 2]

$2y = 1600 \\[0.5em] y = \dfrac{1600}{2} \\[0.5em] \Rightarrow y = 800 m \\[0.5em]$

Therefore, y = 800 m

Substituting the value of y in eqn 1 we get the value of x,

$800 - x = 640 \\[0.5em] \Rightarrow x = 800 - 640 \\[0.5em] \Rightarrow x = 160 \\[0.5em]$

Therefore, x = 160 m

Hence,

x = 160 m and y = 800 m