Free-Fall Acceleration — Question 4

Question 4

A window transit reveals speed but needs a release assumption

A small pointlike object falls vertically past a window of height $h=2.40\ \mathrm{m}$ in $\Delta t=0.200\ \mathrm{s}$. It is already moving downward at the upper edge. Assume free fall with constant $g=10.0\ \mathrm{m/s^2}$ and no air resistance.

1. Use downward positive to find the speeds at the upper and lower edges. Explain why $h/\Delta t$ is neither edge speed.

2. If the object was released from rest above the window, find the distance from its release point to the upper edge and the elapsed time to that edge.

3. Remove the released-from-rest assumption. Give a different release height and a nonzero downward release speed that reproduce the same window transit. Does the transit alone locate the release point?

4. Explain why the window observation cannot distinguish different masses in this ideal model.

Question 4 - Solution

A window transit reveals speed but needs a release assumption

(a) Account for the speed gained inside the window.

If $w$ is the downward entry speed, $h=w\Delta t+g\Delta t^2/2$. Thus $$\boxed{w=12-1=11.0\ \mathrm{m/s}},\qquad \boxed{w_{\mathrm{exit}}=w+g\Delta t=13.0\ \mathrm{m/s}}.$$ The quotient $h/\Delta t=12.0\ \mathrm{m/s}$ is the interval average, equal to the speed at its middle time for constant acceleration.

(b) A specified initial speed supplies the missing condition.

For release from rest, $w^2=2gH$, where $H$ is height above the upper edge. Therefore $$\boxed{H=\frac{11^2}{20}=6.05\ \mathrm{m}},\qquad \boxed{t_{\mathrm{entry}}=w/g=1.10\ \mathrm{s}}.$$

(c) The same entry state can have different earlier origins.

$$H=\frac{w^2-u_0^2}{2g},\quad t_{\mathrm{entry}}=\frac{w-u_0}{g};\qquad 0\leq u_0<w\ \text{gives a family of release states}.$$

For example, release $\boxed{4.80\ \mathrm{m}}$ above the upper edge with downward speed $\boxed{5.00\ \mathrm{m/s}}$ gives $w^2=5^2+2(10)(4.80)=121$. It reaches the upper edge after $(11-5)/10=0.600\ \mathrm{s}$ and produces the identical window transit. The measurement fixes entry speed under the model, but not a unique release point without an initial-speed condition.

(d) Mass is absent from the assumed acceleration law.

All masses with the same entry state and the same $g$ have the same ideal transit. This conclusion presumes negligible air resistance, not equal behavior for arbitrary objects falling through air.