Free-Fall Acceleration — Question 7

Question 7

The sound of impact arrives after the fall ends

A pointlike stone is released from rest at the top of a vertical shaft. A stationary detector at the release level records the impact sound $T=4.25\ \mathrm{s}$ after release. Assume constant $g=10.0\ \mathrm{m/s^2}$, negligible drag on the stone, a vertical sound path back to the detector, and constant sound speed $c=320\ \mathrm{m/s}$. Impact creates the sound immediately; electronic delay is zero.

1. Express the shaft depth $H$ and total measured delay in terms of the fall time $t_f$. Explain why the acoustic stage has a different equation from the stone’s free fall.

2. Solve for the admissible $t_f$ and $H$. Verify that the two positive stage durations sum to the observation.

3. Calculate the depth inferred by incorrectly treating all 4.25 s as fall time. Find its absolute and percentage overestimate relative to the actual depth.

4. Derive an expression for $t_f$ in terms of $T,c,g$ with no subtraction of nearly equal positive quantities. Check its limit as sound speed becomes arbitrarily large.

Question 7 - Solution

The sound of impact arrives after the fall ends

(a) Add durations belonging to distinct physical stages.

The stone falls from rest through $$H=\tfrac12gt_f^2,\qquad T=t_f+H/c=t_f+\frac{g}{2c}t_f^2.$$ The sound travels at the stipulated constant speed, so its travel time is distance divided by $c$. It is not a falling material point with acceleration $g$.

(b) Retain the nonnegative fall-time root.

With the supplied values, $$4.25=t_f+t_f^2/64,\qquad t_f^2+64t_f-272=0.$$ The roots are 4 s and $-68$ s; only $\boxed{t_f=4.00\ \mathrm{s}}$ is physical. Therefore $$\boxed{H=\tfrac12(10)(4)^2=80.0\ \mathrm{m}}.$$ The sound takes $80/320=0.250\ \mathrm{s}$, giving $4.00+0.250=4.25\ \mathrm{s}$ as measured.

(c) A longer assigned fall time overestimates the depth.

$$H_{\mathrm{naive}}=\tfrac12gT^2=90.3125\ \mathrm{m}.$$ The overestimate is $\boxed{10.3125\ \mathrm{m}}$, or $$\boxed{100\frac{H_{\mathrm{naive}}-H}{H}=12.89\%\ \text{approximately}}.$$

(d) Rationalize the positive root.

$$\frac cg(\sqrt{1+2gT/c}-1)\frac{\sqrt{1+2gT/c}+1}{\sqrt{1+2gT/c}+1}=\frac{2T}{\sqrt{1+2gT/c}+1}.$$

The quadratic formula gives $t_f=(c/g)(\sqrt{1+2gT/c}-1)$. Multiplying numerator and denominator by the conjugate yields $$\boxed{t_f=\frac{2T}{1+\sqrt{1+2gT/c}}}.$$ As $c\to\infty$, the denominator tends to 2 and $t_f\to T$. The sound delay then vanishes, recovering the instantaneous-signal idealization.