SOLUTION: I need your help with this problem: Sound travels at 1100 ft./sec. in still air. Linda finds an old well on a farm she has just bought and wants to know how deep it is. She dr

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Question 47500: I need your help with this problem:
Sound travels at 1100 ft./sec. in still air. Linda finds an old well on a farm she has just bought and wants to know how deep it is. She drops a rock from ground level and hears it splash 4.0 seconds later. Correct to the nearest whole foot, how deep is the well?
Answer by venugopalramana(3286) About Me  (Show Source):
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I need your help with this problem:
Sound travels at 1100 ft./sec. in still air. Linda finds an old well on a farm she has just bought and wants to know how deep it is. She drops a rock from ground level and hears it splash 4.0 seconds later. Correct to the nearest whole foot, how deep is the well?
LET THE WELL BE S FT.DEEP
FIRST ROCK HAS TO FALL S FT. TO HIT THE WATER
THE FORMULA FOR FREE FALL DUE TO GRAVITY IS
S=UT+0.5GT^2.....HERE U=0..DROPPED FROM REST...AND G=32 FT/SEC^2
S=0.5*32T^2=16T^2
T=SQRT(S/16)=SQRT(S)*(1/4)
AFTER THE SPLASH SOUND WILL TAKE S/1100 SECS TO REACH THE EAR.
SO TOTAL TIME
=SQRT(S)/4+S/1100=4
LET SQRT(S)=X
X/4+X^2/1100=4
X^2+275X-4400=0
X={-275+(275^2+4*4400)^0.5}/2 = 15.16 FT.
S=X^2=229.94 FT=230 FT.