A person, standing between two parallel hills, fires a gun. He hears the first echo after 1.5s and the second after 2.5s. if the speed of sound is 332m/s. when he will hear the first common echo from both the hills

Let the person P be at a distance x from hill H₁ and y from H₂ as shown In figure. The time interval between the original sound and echoes from H₁ and H₂ will be respectively
${\mathrm{t}}_1=\frac{2\mathrm{x}}{\mathrm{v}}\text{ and }{\mathrm{t}}_2=\frac{2\mathrm{y}}{\mathrm{v}}$

So the distance between the hills, 
$\mathrm{x}+\mathrm{y}=\frac{\mathrm{v}}{2}\left[{\mathrm{t}}_1+{\mathrm{t}}_2\right]=\frac{332}{2}[1.5+2.5]=664\mathrm{m}$
Now as I echo will be from H₁ after time t₁ while II echo from H₂ after time t₂ , III echo will be produced due to reflection of I echo from H₂ or II echo from H₁, Thus we have t₃ = t₁ + t₂ = 1.5 + 2.5 = 4s thus we can state that III echo will be produced after 4s and in it sound from both I and II echoes will reach simultaneously