Two spheres start falling from the same height at the same…
Two spheres start falling from the same height at the same moment. One starts from rest; the other has an initial downward speed $v_0 = 20\ \mathrm{m/s}$. The first sphere lands $\Delta t = 2\ \mathrm{s}$ after the second. (a) From what height were they released? (b) How long did each fall?
Explanation of Method
Two spheres from the same height, one with a head start.
Step 1: Two equations
Let $t_2$ be the fall time of the faster sphere; the slower one falls $t_1 = t_2 + 2$: $h = \tfrac12 g (t_2 + 2)^2$.
\[ h = v_0 t_2 + \tfrac12 g t_2^2. \]
Step 2: Solve for $t_2$
Setting them equal, the $\tfrac12 g t_2^2$ terms cancel: \[2g\,t_2 + 2g = v_0 t_2 \;\Longrightarrow\; t_2 = \frac{2g}{v_0 - 2g} = \frac{19.62}{20 - 19.62} = \frac{19.62}{0.38}.\]
\[t_2 \approx 51.6\ \mathrm{s}, \qquad t_1 \approx 53.6\ \mathrm{s}\]
Step 3: Height
\[ h = \tfrac12 g t_1^2 = 4.905 \cdot 53.63^2. \]
\[h \approx 14\,100\ \mathrm{m} \approx 14\ \mathrm{km}\]
Result
\[\boxed{\,t_2 \approx 51.6\ \mathrm{s}, \qquad t_1 \approx 53.6\ \mathrm{s}\,}\]
\[\boxed{\,h \approx 14\,100\ \mathrm{m} \approx 14\ \mathrm{km}\,}\]
Remark
Note. The result is extremely sensitive to the value of $g$, because $v_0 = 20$ is very close to $2g = 19.62$: with $g = 9.8\ \mathrm{m/s^2}$ one gets $t_2 = 49\ \mathrm{s}$ and $h \approx 12\,700\ \mathrm{m}$. The problem also ignores air resistance, which over 14 km would matter a great deal; treat the answer as an exercise in free-fall equations.