Two men are standing on the same side of a high wall and at the same distance from it such that they are 400m apart .When one man fires a pistol, the other hears the first report 1.2s after seeing the flash and the second report 0.5s after the first.Explain why he hears two reports and calculate the:a.velocity of sound in air b.perpendicular distance of the men from the wall.
Answers
Answer:
(A) The man hears two reports because of Echo.
(B) Velocity of sound in Air= 333.33 ms⁻¹
(C) Perpendicular distance of the men from the wall= 200 m
Explanation:
Assume that two men are A and B
Distance between A and B= 400 m
time for first sound that B report= 1.2 s
time for second sound (echo) that B report= 0.5s
Velocity of sound in air= distance/time= 400/1.2= 333.33 ms⁻¹
Echo time= 1.2+0.5= 1.7s
Echo distannce= Speed of sound in air× Echo time
= 333.33×1.7= 566.661 m
As two sides of the triangle are congruent, one congruent side will be=
= 566.661/2= 283.35 m
In a isosceles right angle triangle two of the angles will be of 45°, as in ΔCBE; therefore perpendicular distance BE will be= sin45°×BC(one congruent side caluclated in ΔABC)= 0.707×283.335
ANS =200 m perpendicular distance of the men from the wall
Hope it will be helpful
Answer:
Explanation:
Velocity of sound in air=distance/time=400/1.2=333.33m/s.
Distance=velocity ×time/2=333.33×time/2
Time= 1.2+0.5=1.7
Distance=333.33×1.7/2=283.33m
Distance for the 2 men each=283.33/2=142m.