Free-Fall Acceleration — Question 6

Question 6

A delayed second release must catch up before the ground

Point A is dropped from rest from height $H=45\ \mathrm{m}$ at $t=0$. From the same height and on the same vertical line, point B is launched downward at speed $u$ after a delay $\tau=1.00\ \mathrm{s}$. Use $g=10\ \mathrm{m/s^2}$, neglect air resistance, and end the motion at the first meeting or first ground contact. Treat objects as points.

1. Use downward distance $z$ from the launch level. At B’s release, find A’s lead and velocity. Derive the separation after B’s release using elapsed time $s=t-\tau$.

2. For $u=15\ \mathrm{m/s}$, find the meeting time, height above ground and both velocities immediately before meeting. Check that the event is in the allowed domain.

3. Derive the strict lower bound on $u$ for a meeting above the ground. Distinguish merely closing the initial gap from closing it soon enough.

4. Interpret the threshold equality and explain why increasing both masses would not change the result in this model.

Question 6 - Solution

A delayed second release must catch up before the ground

(a) Equal accelerations cancel from the relative equation.

At $t=\tau$, A has $z_A=g\tau^2/2=5\ \mathrm{m}$ and downward velocity $g\tau=10\ \mathrm{m/s}$. For $s\geq0$ before contact, $$z_A=\tfrac12g\tau^2+g\tau s+\tfrac12gs^2,\qquad z_B=us+\tfrac12gs^2.$$ Therefore the lead is $$\boxed{z_A-z_B=\tfrac12g\tau^2-(u-g\tau)s}.$$ It shrinks only if $u>g\tau$.

(b) Solve the linear meeting condition, then check height.

For $u=15\ \mathrm{m/s}$, $5-5s=0$, so $s=1\ \mathrm{s}$ and $\boxed{t=2\ \mathrm{s}}$. Both have fallen $z=20\ \mathrm{m}$; the meeting is $\boxed{25\ \mathrm{m}}$ above ground. Immediately beforehand, downward velocities are $$\boxed{w_A=20\ \mathrm{m/s}},\qquad\boxed{w_B=25\ \mathrm{m/s}}.$$ A would reach ground at $T=\sqrt{2H/g}=3\ \mathrm{s}$. The meeting occurs after B’s release and before either point reaches ground.

(c) Require the gap to vanish before A’s ground time.

$$s_{\mathrm{meet}}=\frac{5}{u-10}<2\quad\Longleftrightarrow\quad u>12.5\ \mathrm{m/s}\qquad(u>10\ \mathrm{m/s}).$$

Available chase time is $T-\tau=2\ \mathrm{s}$. Thus $$\frac{g\tau^2/2}{u-g\tau}<T-\tau, \qquad \boxed{u>g\tau+\frac{g\tau^2}{2(T-\tau)}=12.5\ \mathrm{m/s}}.$$ The denominator must be positive. Speeds $10<u<12.5\ \mathrm{m/s}$ close the gap too slowly; their extrapolated meeting is after A’s ground contact and outside the model.

(d) Equality gives contact at the ground, not above it.

At $u=12.5\ \mathrm{m/s}$ the points meet at $t=3\ \mathrm{s}$ and $z=45\ \mathrm{m}$. A strictly airborne meeting needs the strict inequality. Mass does not enter the common gravitational acceleration or the initial-state constraints; changing it alone changes none of these results.