Question

A man standing in front of a vertical cliff fires a gun and hears the echo after 4 seconds. He moves 150 m towards the cliff and again fires the gun. This time, he hears the echo after 3.6 seconds. Find the distance of the cliff from his first position. (assume velocity of sound is not known) 7500 m 750 m 550 m 1500 m

Answer 1

Distance of the cliff from his first position is 1500 m. Option four is correct.

Explanation:

• Given data

        Duration of first echo $\mathbf{(T_{1})=4 \ s}$

        Duration of second echo $\mathbf{(T_{2})=3.6 \ s}$

        Let speed of sound = V

• For first position of man

        Let distance between man and cliff =$\mathbf{X_{1}}$

        Distance between man and cliff $\mathbf{(X_{1})=\frac{VT_{1}}{2}}$     ...1)

• For second position of man

        Let distance between man and cliff =$\mathbf{X_{2}}$

        Distance between man and cliff $\mathbf{(X_{2})=\frac{VT_{2}}{2}}$     ...2)

• It is given that after move 150 m towards cliff

        $\mathbf{X_{1}-X_{2}=150}$            ...3)

        from equation 1),equation 2)  and equation 3)

        $\mathbf{\frac{VT_{1}}{2}-\frac{VT_{2}}{2}=150}$        ...4)

• After putting the value of respective time in equation 4)

        $\mathbf{\frac{4V}{2}-\frac{3.6V}{2}=150}$

        On solving above equation

        Velocity of sound = $\mathbf{V=750 \ \frac{m}{s}}$

• From from equation 1)

        Distance between man and cliff $\mathbf{=X_{1}=\frac{VT_{1}}{2}=\frac{750\times 4}{2}=1500 \ m}$