Q 1- A person standing in front of a vertical cliff fires a gun and hears its echo in 3s - Physics - Sound

Question

Q 1- A person standing in front of a vertical cliff fires a gun and
hears its echo in 3s The speed of sound in air is 340 m/s
(i)- If the person wants to hear the echo 0 5 s earlier, then how
much distance should he move, toward or away from the cliff?
(ii)- Another person stands behind this person, in the same line
with him and the cliff, at a distance of 170 m and fires a gun in
the air What are the consecutive intervals of time at which the
first person hears two sounds?
(iii)- If the speed of sound changes to 350 m/s then how much
distance should the person move towards or away from the cliff, in
order to hear the echo in the same time? (i e in 3 s)?

Debalina Deb · Accepted Answer

Dear student,
v =340 m/st =3sif the distance between cliff and the man =dthen,for echo,v =2dt⇒2d=vt =340×3=1020 m⇒d=10202=510 m(i)if a person wants to hear the echo 0
5 s earlier then t =(3-0 5)=2
5 secthen,2d=2
5×340=850 m⇒d=425 mso the person needs to come towards the cliff by a distance=(510-425)=85 m(ii)when a second person fires a gunthe first person will hear two soundone directly from the firingand other due to echotime required to hear the direct sound =t1=170340=0
5 sectime required to hear the echo =t2=d+d1v+dv=170+510340+510340=2+1
5=3 5 secso interval of time =3 5-0
5 =3 sec(iiii)now v =350 secbut t2-t1=3 secnow,let the first person is at " d" m away from the clifftime required to hear the direct sound =t1=170350=0
486 sectime required to hear the echo =t2=d+d1v+dv=170+d350+d350now,170+d350+d350-0
486=3170+d350+d350=3 486170+d +d=1219 992d=1049 99d =524
99 minitially the man was at 510 mnow it has to be at 524
99 m =525 m (approx)so, it must move away from the cliff  by a distance =525-510=15 m

Regards