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Physics help?
You are standing on the rim of a canyon. You drop a rock and in 7.0 seconds you hear the sound of the rock hitting the bottom of the canyon. How deep is the canyon if sound travels at a constant velocity of 343 m/s?
1 Answer
- billrussell42Lv 72 years agoFavorite Answer
Ans: The 7 seconds is the sum of two times:
1. time for the rock to fall, and that is d = ½gt² or t = √(2d/g)
2. time for the sound to return up the cliff, and that is t = d/343
The total time is
√(2d/9.8) + d/343 = 7
solve for d
√(2d/9.8) = 7 – d/343
square both sides
d/4.9 = 49 + d²/343² – 14d/343
multiply by 343²
24010d = 5764801 + d² – 4802d
d² – 28812d + 5764801 = 0
quadratic equation:
to solve ax² + bx + c = 0
x = [–b ± √(b²–4ac)] / 2a
d = [28812 ± √(28812²–4•5764801)] / 2
d = [28812 ± √(830131344–23059204)] / 2
d = [28812 ± 28409] / 2
d = 201.5 m or 28610 m
taking the second as incorrect,
d = 201.5 m
but check the math.