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प्रश्न
उत्तर
Here, t1 =5s. Let d be the distance of the hill from the man and v be the velocity of sound.
Then t= `(2"d")/"v"`
5 = `(2"d")/"v " or "d" = (5"v")/2` .................................. (1)
When the man moves, through a distance 310 m towards the wall, then d' = d - 310 and t' = 3s .
. ·. 3= `(2 "d" - 310)/"v"`
or, d - 310 = `"v"/2`; .................................... (ii)
Subtracting (ii) from (i)
d - d + 310 = `(5"v")/2 - (3"v")/2 = (2"v")/2`= v
`therefore` v = 310m/s
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