Question

A Man p faces a vertical wall, 660m away from him. A second man Q stands 200 m behind p. when p fires a gun, Q hears the shot and a echo at the time interval of Z. find Z if the speed of sound = 330m/s​

Answer 1

$\underline{\textsf{\textbf{\purple{ \mapsto Given:}}}}$

• P faces a wall which is 660m away from him .
• Q stands 200m behind P .
• P fires a gun, Q hears the shot and a echo at the time interval of Z .
• The speed of sound is 330m/s .

$\underline{\textsf{\textbf{\purple{ \mapsto To\:Find:}}}}$

• The value of Z .

$\underline{\textsf{\textbf{\purple{ \mapsto Concept\:Used:}}}}$

We know that Time = Distance/Speed . So here we will find the distance travelled by sound waves and then we will divide it by speed of sound that is 330m/s , to obtain the time that is z .

$\underline{\textsf{\textbf{\purple{ \mapsto Answer:}}}}$

$\red{\underline{\tt \dashrightarrow Figure:-}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(12,8)\put(6.5,0){\line(0,1){4}}\put(6.5,4){\line(-1,0){0.5}}\put(6,4){\line(0,-1){4}}\put(5.859,4.3){ \sf Wall }\put(0,0){\line(1,0){6.5}}\put(0.2,1){\circle{0.4}}\put(0.1,0.4){\line(0,1){0.4}}\put(0.3,0.8){\line(0, - 1){0.4}}\put(0.3,0.4){\line( - 1, 0){0.2}}\put(0.26,0){\line(0,1){0.4}}\put(0.12,0){\line(0,1){0.4}}\put(0,1.5){ \sf Q }\put(2,1){\circle{0.4}}\put(1.9,0.4){\line(0,1){0.4}}\put(2.1,0.8){\line(0, - 1){0.4}}\put(2.1,0.4){\line( - 1, 0){0.2}}\put(1.94,0){\line(0,1){0.4}}\put(2.09,0){\line(0,1){0.4}}\put(1.8,1.5){ \sf P }\put(0.8, - 0.3){\vector(-1,0){0.8}}\put(0.805,-0.3){ \sf 200m }\put(1.5, - 0.3){\vector(1,0){0.8}}\put(3.9, - 0.3){\vector(-1,0){1.6}}\put(3.95,-0.3){ \sf 660m }\put(4.7, - 0.3){\vector(1,0){1.8}}\end{picture}$

Firstly the sound will travel ( 200m + 660m ) = 860 m . Secondly the sound will travel only 660m since the sound reached P .

Hence the total distance travelled by the sound will be = $\sf 860m+660m=\red{1520m}$

And , the difference will be 1520m - 200m = 1320 m .

Now here ,

• $\tt\green{ Distance= 1320m}$
• $\tt\green{ Speed = 330m/s}$

Hence time will be $\sf\dfrac{1320m}{330ms^{-1}}$ = 4s.

Hence the time taken is 4s.

Answer 2

Given : Man p faces a vertical wall, 660m away from him. A second man Q stands 200 m behind p. when p fires a gun, Q hears the shot and a echo at the time interval of Z

To Find :  Z if the speed of sound = 330m/s​

Solution:

p faces a vertical wall, 660m away from him

Q stands 200 m behind p.

Distance between P & Q = 200 m

Distance between wall & Q = 660 + 200 = 860 m

1st Sound heard after 200 m

Echo heard after   860 + 660 = 1520 m

Difference  = 1520 - 200  = 1320 m

Speed of Sound = 330 m/s

Time interval  = 1320/330

= 4

Z  = 4  sec

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