Two bodies are released at the same time from a height H =…

Two bodies are released at the same time from a height $H = 7900\ \mathrm{m}$. One starts from rest; the other is given an initial downward speed $v_0 = 200\ \mathrm{m/s}$. How does the distance between them change with time? What is it when the second body reaches the ground? (Air resistance is ignored.)

Explanation of Method

Two bodies falling with different initial speeds.

Step 1: Positions

Measuring height from the ground: $y_1 = H - \tfrac12 g t^2$.

\[ y_2 = H - v_0 t - \tfrac12 g t^2. \]

Step 2: Distance between them

\[ d = y_1 - y_2 = v_0 t. \]

\[d(t) = v_0 t = 200t\ \ (\mathrm{m})\]

Gravity accelerates both bodies equally, so it cancels: relative to each other they move uniformly at 200 m/s.

$\text{height } (\mathrm{m})$ $t\ (\mathrm{s})$ $0$ $10$ $20$ $24.6$ $2000$ $4925$ $7900$ $d = v_0 t \approx 4925\ \mathrm{m}$ $200 \cdot 12 = 2400\ \mathrm{m}$ $\text{body 1 (from rest)}$ $\text{body 2 } (v_0 = 200\ \mathrm{m/s})$

Heights of both bodies against time. Both curves bend by exactly the same amount (the same $g$), so the vertical gap between them grows in a straight line, 200 m every second.

Step 3: When the second body lands

\[ \tfrac12 g t^2 + v_0 t - H = 0 \;\Longrightarrow\; t = \frac{-v_0 + \sqrt{v_0^2 + 2gH}}{g} = \frac{-200 + 441.6}{9.81} \approx 24.63\ \mathrm{s}. \]

\[ d = 200 \cdot 24.63. \]

\[d \approx 4925\ \mathrm{m} \approx 4.9\ \mathrm{km}\]

Check

Body 1 at that moment: $ 7900 - 4.905 \cdot 24.63^2 \approx 7900 - 2975 = 4925\ \mathrm{m} $ above the ground $\checkmark$ (it lands only after $ \sqrt{2H/g} \approx 40\ \mathrm{s} $)

Result

\[\boxed{\,d(t) = v_0 t = 200t\ \ (\mathrm{m})\,}\]

\[\boxed{\,d \approx 4925\ \mathrm{m} \approx 4.9\ \mathrm{km}\,}\]