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 initial speed $v_0$. What must $v_0$ be for the bodies to meet halfway, at $H/2$?

Step 1: When does the falling body reach the middle?

\[ \frac H2 = \tfrac12 g t^2 \;\Longrightarrow\; t = \sqrt{\frac{H}{g}} = \sqrt{\frac{8000}{9.81}} \approx 28.56\ \mathrm{s}. \]

Step 2: Use relative motion

Both bodies have the same acceleration $g$, so it cancels between them: they approach each other at the constant speed $v_0$. They meet when they have closed the whole gap $H$:

\[ v_0 t = H \;\Longrightarrow\; v_0 = \frac{H}{t} = \frac{8000}{28.56}. \]

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

Check

Rising body at 28.56 s: $ 280.1 \cdot 28.56 - 4.905 \cdot 28.56^2 = 8000 - 4000 = 4000\ \mathrm{m}\ \checkmark $

Its speed then: $ 280.1 - 9.81 \cdot 28.56 \approx 0 $: it meets the other body exactly at the top of its flight