Free-Fall Acceleration — Question 10

Question 10

Guarantee a descending arrival when gravity is only bounded

A pointlike payload is launched straight upward from $y=0$ with an exactly controlled speed $u>0$. A receiver at height $h=15.0\ \mathrm{m}$ remains inactive on ascent and makes first contact only on descent. It requires a strictly downward arrival with speed no greater than $V=8.00\ \mathrm{m/s}$. The path is otherwise unobstructed.

The unknown local gravitational acceleration is constant during the flight, with a hard bound $9.50\leq g\leq10.50\ \mathrm{m/s^2}$. Neglect drag and all contact interactions before reception. These are possible values, not statistical uncertainties.

1. For a fixed $g$, express the descending arrival speed in terms of $u,g,h$. Explain why merely reaching the receiver height at the highest point fails the strictly downward requirement.

2. Find the complete interval of launch speeds that guarantees both requirements for every allowed $g$. Identify the worst value of $g$ for each constraint and retain the correct open or closed endpoints.

3. Test $u=18.0\ \mathrm{m/s}$. Give the full possible range of descending arrival speeds and verify the guarantee without assuming a midpoint value of $g$.

4. Among all guaranteed launch speeds, determine the largest possible apex height above launch level and identify values that attain it. Keep the analysis within the stated ideal free-fall model.

Question 10 - Solution

Guarantee a descending arrival when gravity is only bounded

(a) Use the descending branch and require a genuine crossing.

$$t_{\mathrm{desc}}=\frac{u+\sqrt{u^2-2gh}}g,\qquad v(t_{\mathrm{desc}})=-\sqrt{u^2-2gh}<0.$$

At height $h$, $v^2=u^2-2gh$. On descent, $v=-\sqrt{u^2-2gh}$, so arrival speed is $w=\sqrt{u^2-2gh}$. A strictly downward crossing requires $u^2>2gh$. Equality places the receiver at the apex, where velocity is zero; it is not a descending arrival.

(b) Different constraints have different worst-case gravities.

To pass above the receiver even at the largest gravity requires $u^2>2g_{\max}h=315\ \mathrm{m^2/s^2}$. Arrival speed is largest at the smallest gravity, so the upper-speed condition requires $$u^2-2g_{\min}h\leq V^2,\qquad u^2\leq285+64=349\ \mathrm{m^2/s^2}.$$ Consequently the complete feasible interval is $$\boxed{\sqrt{315}<u\leq\sqrt{349}\quad\text{in m/s}},$$ approximately $\boxed{17.748<u\leq18.682\ \mathrm{m/s}}$. The lower endpoint is excluded by the strict descent condition; the upper endpoint is allowed because 8 m/s is permitted.

(c) Evaluate both attainable extreme speeds.

For $u=18.0\ \mathrm{m/s}$, $$w_{\min}=\sqrt{324-315}=3.00\ \mathrm{m/s},\qquad w_{\max}=\sqrt{324-285}=\sqrt{39}\approx6.245\ \mathrm{m/s}.$$ Thus $\boxed{3.00\leq w\leq6.245\ \mathrm{m/s}\ \text{approximately}}$, always positive and below 8 m/s. The signed velocities are negative throughout this descending-arrival range.

(d) Maximize height with respect to both allowed parameters.

Apex height $u^2/(2g)$ increases with $u^2$ and decreases with $g$. Its largest possible value is attained at the allowed upper launch speed and lower gravity: $$\boxed{y_{\max,\,largest}=\frac{349}{2(9.50)}=\frac{349}{19}\approx18.368\ \mathrm{m}}.$$ At these same parameters the descending arrival speed is exactly 8 m/s, so the maximum is attainable within the guarantee, not just a limiting value.