(a) A man fires a gun and hears its echo after 3 s. The man then moves 80 m towards the hill and fires his gun again. This time he hears the echo after 2.5 s. Calculate the speed of the sound.

(b) State one reason of using ultrasonic waves in SONAR.

Given,

• Time to hear first echo (t₁) = 3 s
• Time to hear second echo (t₂) = 2.5 s
• Distance moved by the man towards the hill (d) = 80 m

(a) Let original distance of the man from the cliff is 'D' and speed of sound be 'v'.

Case 1 : When person hears first echo from initial position.

$\text {Distance of man from cliff} = \dfrac {\text {Speed of sound} \times \text {Time to hear first echo}}{2}\\[1em] \Rightarrow \text D = \dfrac {\text v \times \text t_1}{2} = \dfrac{3\text v}{2}$

Case 2 : When person hears second echo after moving 80 m towards the cliff.

New distance between the man and the cliff = D - d = $\dfrac{3\text v}{2}$ - 80

Then,

$\text {Distance of the man from the cliff} = \dfrac {\text {Speed of sound} \times \text {Time to hear second echo}}{2}\\[1em] \Rightarrow \dfrac {3\text v}{2} - 80 = \dfrac {\text v \times \text t_2}{2} \\[1em] \Rightarrow \dfrac {3\text v}{2} - 80 = \dfrac{2.5\text v}{2} \\[1em] \Rightarrow \dfrac {3\text v}{2} - 80 = \dfrac{25\text v}{20} \\[1em] \Rightarrow \dfrac {3\text v}{2} - 80 = \dfrac{5\text v}{4} \\[1em] \Rightarrow \dfrac {3\text v}{2} - \dfrac{5\text v}{4} = 80 \\[1em] \Rightarrow \dfrac {6\text v - 5\text v}{4} = 80 \\[1em] \Rightarrow \dfrac {\text v}{4} = 80$

⇒ v = 80 x 4 = 320 m s^-1

Hence, speed of the sound is 320 m s^-1.

(b) Ultrasonic waves can travel undeviated through a long distance and that's why they are used in SONAR.