a man standing in front of a large wall , claps two objects against each other at an interval of 1.2 seconds regularly. the echo of the first clap coincides with the fifth clap. if the speed of sound in air is 340 m/s the distance between the man and the wall is ?

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Solution

The fifth clap happens $1.2 \times 4 = 4.8 s e c$ after the first clap. In this time interval the sound travels $340 \times 4.8 = 1632 m$ ,
So the wall is $\frac{1632}{2} = 816 m$ away from the man.

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a man standing in front of a large wall , claps two objects against each other at an interval of 1.2 seconds regularly. the echo of the first clap coincides with the fifth clap. if the speed of sound in air is 340 m/s the distance between the man and the wall is ?

The fifth clap happens 1.2×4=4.8 sec after the first clap. In this time interval the sound travels 340×4.8=1632 m, So the wall is 16322=816 m away from the man.

The fifth clap happens $1.2 \times 4 = 4.8 s e c$ after the first clap. In this time interval the sound travels $340 \times 4.8 = 1632 m$ ,
So the wall is $\frac{1632}{2} = 816 m$ away from the man.