The correct option is C 310ms−1
The speed of the sound is calculated as follows.
Let d be the distance between the man and the hill in the beginning.
v=2×dt = 2×d5→eqn1
He moves 310 m towards the hill. Therefore the distance will be (d - 310) m. Therefore,
v=2(d−310)3→eqn2
Since velocity of sound is same, equating (1) and (2), we get
2d5=2(d−310)3
3d = 5d - 1550
2d = 1550
d = 775 m
Hence, the velocity of sound v=2×7755 (substituting in equation 1)
v=310m/s