Problem Statement
A body falls from height $h$ with air drag $F_{drag} = \alpha v$. It reaches speed $v_0$ just before impact. Find $h$ (qualitative approach since terminal velocity is relevant).
Given Information
- See problem statement for all given quantities.
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: This requires solving the ODE $m\dot v = mg – \alpha v$.
Step 2 — Apply the relevant physical law or equation: Terminal velocity: $v_t = mg/\alpha$
Step 3 — Solve algebraically for the unknown: Solution: $v(t) = v_t(1-e^{-\alpha t/m})$
Step 4 — Substitute numerical values with units: Position: $y(t) = v_t t + \frac{m v_t}{\alpha}(e^{-\alpha t/m}-1) = v_t t – \frac{m}{\alpha}(v_t – v(t)) \cdot \frac{m}{\alpha}$
Step 5 — Compute and check the result: Energy approach — work-energy theorem:
Step 6: $$mgh – W_{drag} = \frac12 mv_0^2$$
$$W_{drag} = \int_0^h \alpha v\,ds$$
Worked Calculation
$$mgh – W_{drag} = \frac12 mv_0^2$$
$$W_{drag} = \int_0^h \alpha v\,ds$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
This requires solving the ODE $m\dot v = mg – \alpha v$.
Terminal velocity: $v_t = mg/\alpha$
Solution: $v(t) = v_t(1-e^{-\alpha t/m})$
Position: $y(t) = v_t t + \frac{m v_t}{\alpha}(e^{-\alpha t/m}-1) = v_t t – \frac{m}{\alpha}(v_t – v(t)) \cdot \frac{m}{\alpha}$
Energy approach — work-energy theorem:
$$mgh – W_{drag} = \frac12 mv_0^2$$
$$W_{drag} = \int_0^h \alpha v\,ds$$
This is complex without knowing $v(s)$. For the simpler case of drag proportional to $v^2$, quadrature gives $h = \frac{v_t^2}{2g}\ln\frac{v_t^2}{v_t^2-v_0^2}$ where $v_t = \sqrt{mg/k}$.
Answer
$$\boxed{W_{drag} = \int_0^h \alpha v\,ds}$$
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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