Free-Fall Acceleration — Question 8

Question 8

A last-second observation can create an inadmissible root

A point is dropped from rest and falls to the ground under constant $g=10\ \mathrm{m/s^2}$ with negligible air resistance. During the last $\Delta=1.00\ \mathrm{s}$ before impact, it travels exactly $f=5/9$ of the total drop height. The whole flight lasts at least one second.

1. Express the last-second distance as the difference between two positions, using total fall duration $T$. Derive the quadratic equation for $T$.

2. Find both algebraic roots and select the admissible one. Determine the total height and the distances traveled before and during the last second.

3. Find speed at the beginning of the last second and immediately before impact. Verify the last-second distance using its average speed.

4. For general $0<f<1$ and a last interval of length $\Delta$, derive the physically admissible expression for $T$. Explain why the other branch corresponds to extending the trajectory to a time before release.

Question 8 - Solution

A last-second observation can create an inadmissible root

(a) Subtract distances measured from the same release event.

The total height is $H=gT^2/2$. Because $T\geq\Delta$, the final interval starts at a valid elapsed time $T-\Delta$. Its distance is $$D_{\mathrm{last}}=\tfrac12g[T^2-(T-\Delta)^2] =gT\Delta-\tfrac12g\Delta^2.$$ Setting this equal to $fH$ gives $fT^2-2\Delta T+\Delta^2=0$. For the stated values, $\boxed{5T^2-18T+9=0}$ with numerical seconds.

(b) The time domain rejects an otherwise valid algebraic root.

$$5T^2-18T+9=(5T-3)(T-3)=0;\qquad T\geq1\ \mathrm{s}\Rightarrow T=3\ \mathrm{s}.$$

The roots are $T=3.00\ \mathrm{s}$ and $T=0.600\ \mathrm{s}$. The latter is shorter than the stipulated last second, so it is inadmissible. Thus $$\boxed{T=3.00\ \mathrm{s}},\qquad\boxed{H=45.0\ \mathrm{m}}.$$ By $T-\Delta=2\ \mathrm{s}$, the point has fallen 20 m; during the last second it falls 25 m. Indeed $25/45=5/9$.

(c) Check against the interval’s actual endpoint speeds.

The downward speeds are $\boxed{20\ \mathrm{m/s}}$ at 2 s and $\boxed{30\ \mathrm{m/s}}$ at 3 s. Their mean is 25 m/s, so the last-second distance is $(25\ \mathrm{m/s})(1\ \mathrm{s})=25\ \mathrm{m}$.

(d) Apply the domain before choosing a square-root branch.

Since $f=1-(1-\Delta/T)^2$ and $T\geq\Delta$, the quantity $1-\Delta/T$ is nonnegative. Therefore $$\boxed{T=\frac{\Delta}{1-\sqrt{1-f}}} =\frac{\Delta(1+\sqrt{1-f})}{f}.$$ The other root is $\Delta/(1+\sqrt{1-f})<\Delta$. It would evaluate the position at $T-\Delta<0$, before release. An algebraic continuation of the parabola does not make that earlier interval part of the specified fall.