Question

Two bombs were exploded at a place P with a time interval of 40 minutes. A person moving away from P heard the first explosion at a point A and the second explosion when he was at a point B. If he heard the explosions at an interval of 41 minutes and the speed of sound is 331 m/s, what is the distance between A and B (in m)?​

Answer 1

Answer:

If

Step-by-step explanation:

If we use the formula

s=ut+1/2gt square

T1=40min

T2=41min

V=331m/s

g=10m/s square

S=1/2 into 10 into 1 (u=0,t2-t1=t)

Then S=5m

Answer 2

Answer:

Distance between A and B is 19860 m

Step-by-step explanation:

Given

speed of sound is 331 m/s

time taken to heard the explosion=(t2-t1)=41-40=1 min=60 sec

Distance between A and B =S*t

                                             =331*60

                                              =19,860 m

Distance between A and B is 19860 m

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