Problem Statement
A cannon fires a shell with initial velocity $v_0 = 240\,\text{m/s}$. Target is $l = 5.10\,\text{km}$ away at the same level. Find possible flight times (no air drag).
Given Information
- $v_0 = 240\,\text{m/s}$
- $l = 5.10\,\text{km}$
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: Range formula: $l = v_0^2\sin2\alpha/g$
Step 2 — Apply the relevant physical law or equation: $$\sin2\alpha = \frac{lg}{v_0^2} = \frac{5100\times9.8}{240^2} = \frac{49980}{57600} \approx 0.868$$
$$2\alpha_1 \approx 60.1°\implies\alpha_1\approx30°; \quad 2\alpha_2\approx119.9°\implies\alpha_2\approx60°$$
Step 3 — Solve algebraically for the unknown: Flight times $T=2v_0\sin\alpha/g$:
Step 4 — Substitute numerical values with units: $$T_1 = \frac{2\times240\times\sin30°}{9.8} = \frac{240}{9.8} \approx \boxed{24\,\text{s}}$$
$$T_2 = \frac{2\times240\times\sin60°}{9.8} = \frac{240\sqrt{3}}{9.8} \approx \boxed{42\,\text{s}}$$
Worked Calculation
$$\sin2\alpha = \frac{lg}{v_0^2} = \frac{5100\times9.8}{240^2} = \frac{49980}{57600} \approx 0.868$$
$$2\alpha_1 \approx 60.1°\implies\alpha_1\approx30°; \quad 2\alpha_2\approx119.9°\implies\alpha_2\approx60°$$
$$T_1 = \frac{2\times240\times\sin30°}{9.8} = \frac{240}{9.8} \approx \boxed{24\,\text{s}}$$
Range formula: $l = v_0^2\sin2\alpha/g$
$$\sin2\alpha = \frac{lg}{v_0^2} = \frac{5100\times9.8}{240^2} = \frac{49980}{57600} \approx 0.868$$
$$2\alpha_1 \approx 60.1°\implies\alpha_1\approx30°; \quad 2\alpha_2\approx119.9°\implies\alpha_2\approx60°$$
Flight times $T=2v_0\sin\alpha/g$:
$$T_1 = \frac{2\times240\times\sin30°}{9.8} = \frac{240}{9.8} \approx \boxed{24\,\text{s}}$$
$$T_2 = \frac{2\times240\times\sin60°}{9.8} = \frac{240\sqrt{3}}{9.8} \approx \boxed{42\,\text{s}}$$
Answer
$$T_1 = \frac{2\times240\times\sin30°}{9.8} = \frac{240}{9.8} \approx \boxed{24\,\text{s}}$$
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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