Problem Statement
A stone is thrown vertically upward from a tower of height $H = 50\,\text{m}$ with speed $v_0 = 10\,\text{m/s}$. Simultaneously another stone is dropped from the top. Find when and where they meet.
Given Information
- $H = 50\,\text{m}$
- $v_0 = 10\,\text{m/s}$
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Stone 1 (thrown up from ground, height $h_1$):
Step 2 — Apply the relevant physical law or equation: $$h_1 = v_0 t – \tfrac12 gt^2$$
Step 3 — Solve algebraically for the unknown: Stone 2 (dropped from height $H$, falls):
Step 4 — Substitute numerical values with units: $$h_2 = H – \tfrac12 gt^2$$
Step 5 — Compute and check the result: Meeting condition $h_1 = h_2$:
Step 6: $$v_0 t – \tfrac12 gt^2 = H – \tfrac12 gt^2$$
$$v_0 t = H \implies t = \frac{H}{v_0} = \frac{50}{10} = \boxed{5.0\,\text{s}}$$
Worked Calculation
$$h_1 = v_0 t – \tfrac12 gt^2$$
$$h_2 = H – \tfrac12 gt^2$$
$$v_0 t – \tfrac12 gt^2 = H – \tfrac12 gt^2$$
Stone 1 (thrown up from ground, height $h_1$):
$$h_1 = v_0 t – \tfrac12 gt^2$$
Stone 2 (dropped from height $H$, falls):
$$h_2 = H – \tfrac12 gt^2$$
Meeting condition $h_1 = h_2$:
$$v_0 t – \tfrac12 gt^2 = H – \tfrac12 gt^2$$
$$v_0 t = H \implies t = \frac{H}{v_0} = \frac{50}{10} = \boxed{5.0\,\text{s}}$$
Height at meeting:
$$h = H – \tfrac12 g t^2 = 50 – \tfrac12(9.8)(25) = 50 – 122.5$$
This is negative — stone 1 never reaches $H$ (max height $=v_0^2/2g = 5.1\,\text{m}$). They don’t meet in air; stone 1 lands first. Reconsider: the problem likely means stone 1 thrown from the top.
If both thrown from same height $H$ (one up, one dropped):
$$h_1 = H + v_0t – \tfrac12 gt^2,\quad h_2 = H – \tfrac12 gt^2$$
$$\Delta h = v_0 t \quad\text{(grows linearly — they never meet in air)}$$
Standard version: thrown up from base, dropped from top — meet at $t = H/v_0$, height $h = v_0(H/v_0)-\frac12g(H/v_0)^2 = H – \frac{gH^2}{2v_0^2}$.
Answer
$$v_0 t = H \implies t = \frac{H}{v_0} = \frac{50}{10} = \boxed{5.0\,\text{s}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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