Asked by Larry
Drop a stone in a well, the sound is heard 17 seconds later. How deep is the well in feet
Answers
Answered by
Reiny
d = -16t^2 , if we have to consider direction
or
d = 16t^2 if all we care about is distance
when t = 17
d = 16(17)^2 = 4624 ft
some deep well !!
or
d = 16t^2 if all we care about is distance
when t = 17
d = 16(17)^2 = 4624 ft
some deep well !!
Answered by
Henry
T1 = Time for stone to hit bottom.
T2 = Time for sound to reach top of well
T1 + T2 = 17 s.
T1 = 17-T2
0.5g*(17-T2)^2 = Vs*T2
4.9(289-34T2+T2^2) = 343T2
1416-167T2+4.9T2^2 = 343T2
4.9T2^2-510T2+1416 = 0
Use Quadratic formula and get:
T2 = 2.85 s.
T1 = 17-2.85 = 14.15 s.
d = Vs*T2 = 343 * 2.85 = 978 m = 3226 Ft
T2 = Time for sound to reach top of well
T1 + T2 = 17 s.
T1 = 17-T2
0.5g*(17-T2)^2 = Vs*T2
4.9(289-34T2+T2^2) = 343T2
1416-167T2+4.9T2^2 = 343T2
4.9T2^2-510T2+1416 = 0
Use Quadratic formula and get:
T2 = 2.85 s.
T1 = 17-2.85 = 14.15 s.
d = Vs*T2 = 343 * 2.85 = 978 m = 3226 Ft
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