Free-Fall Acceleration — Question 5

Question 5

Two crossings can measure an unknown local gravity

In a vacuum chamber at an unspecified location, a pointlike ball is launched vertically upward from $y=0$ at $t=0$ with unknown speed $u$. It passes a noncontact sensor at $y=12\ \mathrm{m}$ at $t_1=1.00\ \mathrm{s}$ on ascent and $t_2=3.00\ \mathrm{s}$ on descent. Its downward acceleration has unknown constant magnitude $g$. Do not assume an Earth value.

1. Use both observations to determine $u$ and $g$. Verify the direction at each crossing.

2. Derive general expressions for $g$ and $u$ in terms of a positive sensor height $h$ and two positive crossing times $t_1<t_2$ after launch from zero height.

3. Find the apex time and height and the return time to launch level. Would a sensor at 18 m record a passage? How many distinct passage times would a sensor exactly at the apex height have?

4. Explain which symmetry makes the apex time the mean of the two crossing times, and why the entire interval between those times is still free fall.

Question 5 - Solution

Two crossings can measure an unknown local gravity

(a) Fit a single trajectory to both observations.

From $y=ut-gt^2/2$, $$12=u-\tfrac12g,\qquad12=3u-\tfrac92g.$$ Eliminating $u$ gives $24=3g$, hence $$\boxed{g=8.00\ \mathrm{m/s^2}},\qquad\boxed{u=16.0\ \mathrm{m/s}}.$$ Velocity is $v=16-8t$: at 1 s it is $+8\ \mathrm{m/s}$, and at 3 s it is $-8\ \mathrm{m/s}$, as required.

(b) Use the two roots of the level-crossing equation.

The equation is $gt^2/2-ut+h=0$, so $$\frac g2(t-t_1)(t-t_2)=\frac g2t^2-ut+h.$$ Equating constant and linear coefficients gives $$\boxed{g=\frac{2h}{t_1t_2}},\qquad \boxed{u=\frac g2(t_1+t_2)=\frac{h(t_1+t_2)}{t_1t_2}}.$$ These formulas depend on a known launch time, launch height zero, and one constant $g$ throughout.

(c) Test reachability before solving for passage times.

$$\boxed{t_{\mathrm{top}}=u/g=2.00\ \mathrm{s}},\qquad \boxed{y_{\mathrm{top}}=u^2/(2g)=16.0\ \mathrm{m}}.$$ The return to zero height occurs at $2u/g=4.00\ \mathrm{s}$. The 18 m level is unreachable. At 16 m the two roots coincide at 2 s: there is one distinct instant of contact with that level and no crossing from one side to the other.

(d) Equal-height passages straddle the highest point.

$$y(t)=16-4(t-2)^2\quad\text{(SI)},\qquad y(2-\delta)=y(2+\delta),\quad v(2-\delta)=-v(2+\delta).$$

Completing the square yields $y=y_{\mathrm{top}}-g(t-u/g)^2/2$. Equal heights have times equally spaced around $u/g$, so $t_{\mathrm{top}}=(t_1+t_2)/2$. Gravity acts throughout both branches and at the instant of zero velocity; upward motion is also free fall.