A stone is dropped from rest into a well. The sound of the splash is heard 2.5 s after the stone is released. Assuming the speed of sound is 340 m/s and neglecting air resistance, what is the approximate depth of the well? (Take g = 9.8 m/s² and use the quadratic formula if needed.)
Show the answer and explanation
Correct answer
A. 28 m
Principle or equation
The stone falls under gravity with constant acceleration g. The time to fall is t1, and the sound travels back up in time t2, with t1 + t2 = 2.5 s. The depth d satisfies d = (1/2)g t1² and d = v_sound t2.
Why this answer is correct
Let t1 be the fall time and t2 = 2.5 - t1 the sound return time. Since d = (1/2)g t1² = v_sound t2, we have 4.9 t1² = 340(2.5 - t1). Rearranging gives 4.9 t1² + 340 t1 - 850 = 0. Solving: t1 = [-340 + sqrt(340² + 4*4.9*850)]/(2*4.9) ≈ 2.39 s. Then d = 4.9*(2.39)² ≈ 28 m.
Example
If total time were 2.0 s and sound speed 340 m/s, the equation would be 4.9 t1² = 340(2 - t1), giving t1 ≈ 1.94 s and d ≈ 18.4 m.
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