Problem Statement
Solve the kinematics problem: A projectile is fired at 20 m/s at 60° above horizontal. Find the average acceleration over the first 2 s. ($g = 10$ m/s²) $\vec{a}_{avg} = \Delta\vec{v}/\Delta t$; for projectile under gravity, $\vec{a} = -g\hat{j}$ always Step 1: The acceleration is constant: $\vec{a} = -10\hat{j}$ m/s² at all tim
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Projectile motion decomposes into independent horizontal and vertical components. Horizontal: constant velocity (no air resistance). Vertical: constant downward acceleration $g$. The trajectory is a parabola. Maximum range occurs at $45°$ launch angle; max height at $90°$.
- $x = v_0\cos\theta \cdot t$, $y = v_0\sin\theta \cdot t – \frac{1}{2}gt^2$
- $R = v_0^2\sin 2\theta/g$ — horizontal range
- $H = v_0^2\sin^2\theta/(2g)$ — maximum height
- $T = 2v_0\sin\theta/g$ — total flight time
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$R = \frac{u^2\sin 2\theta}{g} = \frac{400\times\sin 60°}{9.8} = \frac{400\times0.866}{9.8} = \frac{346.4}{9.8} \approx 35.3\,\text{m}$$
$$H = \frac{u^2\sin^2\theta}{2g} = \frac{400\times0.25}{19.6} = \frac{100}{19.6} \approx 5.1\,\text{m}$$
$$\boxed{R = \dfrac{u^2\sin 2\theta}{g},\quad H = \dfrac{u^2\sin^2\theta}{2g}}$$
Answer
$$\boxed{R = \dfrac{u^2\sin 2\theta}{g},\quad H = \dfrac{u^2\sin^2\theta}{2g}}$$
Physical Interpretation
The trajectory is a parabola because gravity provides constant downward acceleration while horizontal velocity remains constant (absent air resistance). Real projectiles deviate due to drag, especially at high speeds.
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