A man drops a rock into a well. (a) the man hears the sound of the splash 2.40 s after he releases the rock from rest. the speed
of sound in air (at the ambient temperature) is 336 m/s. how far below the top of the well is the surface of the water? (b) what if? if the travel time for the sound is ignored, what percentage error is introduced when the depth of the well is calculated?
Part (a) Let h = depth of water surface from the top of the well, m
Because the stone is dropped, its initial velocity is zero. Let t₁ = time for the stone to drop, s Then h = (1/2)*(9.8 m/s²)*(t₁ s)² t₁ = √(h/4.9) s = 0.4518√h s
The velocity of sound is 336 m/s. The time for the sound wave to travel from the water surface to the top of the well is t₂ = h/336 m/s
Because the time before the sound is heard is 2.4 s, therefore t₁ + t₂ = 2.4 0.45181√h + h/336 = 2.4 Multiply through by 336. 151.8082√h + h = 806.4 h - 806.4 = -151.8082√h h² - 1612.8h + 6.5028 x 10⁵ = 2.3046 x 10⁴h h² - 2.4659 x 10⁴h + 6.5028 x 10⁵ = 0
Solve with the quadratic formula. h = 0.5[2.4659 x 10⁴ +/- 2.4606 x 10⁴] = 24632 m or 26.5 m
Test the two answers. When h = 24632, t₁ = 0.4518√(24632) = 70.9 s (Not acceptable) When h = 26.5 m, t₁ = 0.4518√(26.5) = 2.3258 s, t₂ = 26.5/336 = 0.0789 s, t₁+t₂ = 2.4 s (CORRECT)
Answer: h = 26.5 m
Part (b) If travel time for the sound is ignored, then 0.4518√h = 2.4 h = (2.4/0.45118)² = 28.2 m The percent error is 100*[(28.2 - 26.5)/26.5] = 6.42%
The amount of energy it takes to lift a box might be a function of the weight of the objects inside the box. Work is proportional to force and distance. The force of the box is the weight itself of the box. Hence the answer to this problem is B.
