One body is released from rest at a height H = 8000 m . At…

One body is released from rest at a height $H = 8000\ \mathrm{m}$. At the same moment another body is thrown vertically upward from the ground along the same line with speed $v_0$. What must $v_0$ be for the bodies to meet at $H/2$? What is the velocity of the first body relative to the second when they meet?

Explanation of Method

Meeting halfway and the relative velocity at the meeting.

Step 1: Initial speed

As in Gravitational Kinematics 007: the falling body reaches $H/2$ at $t = \sqrt{H/g} \approx 28.56\ \mathrm{s}$, and the bodies close the gap $H$ at the constant relative speed $v_0$:

\[v_0 = \frac{H}{t} = \sqrt{gH} \approx 280\ \mathrm{m/s}\]

Step 2: Velocities at the meeting

Taking upward as positive: $v_1 = -g t \approx -280\ \mathrm{m/s}$.

\[ v_2 = v_0 - g t \approx 0. \]

Step 3: Relative velocity

\[ \vec v_{12} = \vec v_1 - \vec v_2 = (-gt) - (v_0 - gt) = -v_0. \]

\[v_{\text{rel}} = v_0 \approx 280\ \mathrm{m/s}, \ \text{directed downward}\]

$v\ (\mathrm{m/s}, \text{ up} +)$ $t\ (\mathrm{s})$ $280$ $v_{\text{rel}} = 280\ \mathrm{m/s}$ $280$ $28.6$ $\text{falling body}$ $\text{rising body}$

falling body
rising body

Velocities against time (upward positive). Both lines have the same slope $-g$, so they stay 280 m/s apart at all times: the relative velocity never changes. At the meeting, the rising body is at rest and the falling one moves at 280 m/s.

Result

\[\boxed{\,v_0 = \frac{H}{t} = \sqrt{gH} \approx 280\ \mathrm{m/s}\,}\]

\[\boxed{\,v_{\text{rel}} = v_0 \approx 280\ \mathrm{m/s}, \ \text{directed downward}\,}\]

Remark

Note. The relative velocity is $-v_0$ at every moment, not only at the meeting, because the same $g$ acts on both bodies. At the meeting it happens that the rising body is momentarily at rest, so the whole 280 m/s belongs to the falling body.