JAMB 1980
A man stands one-third of the way between two walls and claps his hands, He hears two distinct echoes 1s apart. If the speed of sound is 330ms-1 the distance between the wall is
A. 33m
B. 110m
C. 220m
D. 330m
E. 495m
Correct Answer: Option E
Explanation
the man was standing at one third of d(distance between two walls), that means that he is 2/3 d from the other wall, and the echo were distinct echoes (not same reflection)
so since the v is same, make v subject for the two echoes
first echo distance=1/3d, time =x,
for second echo distance=2/3d, time =(x+1)seconds
from eqn v=2d/t
therefore (2(1/3d))/x =(2(2/3d))/x+1
after solving this u would get x =1
then sub 1 for x in the equation v=2d/t to get ur d, so 330=2(1/3d)/1, d =495
Or
v=2d/t
but the man stands one third of his distance =⅓d
Therefore, v=2×⅓d/t
330=2×⅓d/1
330=2d/3
Making d the subject formula
2d=330×2
d=990/2
d=495
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Sir how did we get the Time
Type here..sir i dont thinl the answer is correct the answer should be 330 not 495