Two bodies, A and B, fall freely from rest. Body B falls…

Two bodies, A and B, fall freely from rest. Body B falls from a height $h_B = 150\ \mathrm{m}$ and lands $\Delta t = 3.5\ \mathrm{s}$ earlier than body A. From what height was body A released?

Explanation of Method

Two bodies falling from different heights.

Step 1: Fall time of B

\[ h_B = \tfrac12 g t_B^2 \;\Longrightarrow\; t_B = \sqrt{\frac{2h_B}{g}} = \sqrt{\frac{300}{9.81}} \approx 5.53\ \mathrm{s}. \]

Step 2: Fall time of A

\[ t_A = t_B + \Delta t = 5.53 + 3.5 = 9.03\ \mathrm{s}. \]

Step 3: Height of A

\[ h_A = \tfrac12 g t_A^2 = 4.905 \cdot 9.03^2. \]

\[h_A \approx 400\ \mathrm{m}\]

$\text{distance fallen } (\mathrm{m})$ $t\ (\mathrm{s})$ $0$ $2$ $4$ $5.53$ $9.03$ $150$ $400$ $\text{B lands}$ $\text{A lands}$ $\Delta t = 3.5\ \mathrm{s}$ $h = \tfrac12 g t^2$

Both bodies follow the same free-fall curve; they just stop at different heights. B lands after 5.53 s, A needs 3.5 s more, and in that extra time it falls much further because it is already fast.

Result

\[\boxed{\,h_A \approx 400\ \mathrm{m}\,}\]

Remark

Note. The heights are not in the ratio of the times (9.03/5.53 ≈ 1.6) but of their squares (≈ 2.7): $400/150 \approx 2.7$. That is because the distance fallen grows with $t^2$.