Problem Statement
Solve the kinematics problem: A ball is thrown at speed $u = 20$ m/s at angle $45°$ with the horizontal. Find the horizontal and vertical components of velocity. $v_x = u\cos\theta$, $v_y = u\sin\theta$ Step 1: $v_x = 20\cos 45° = 20/\sqrt{2} = 10\sqrt{2} \approx 14.14$ m/s. Step 2: $v_y = 20\sin 45° = 10\sqrt{2} \approx 14.14$
Given Information
- $v_x = 20\cos 45° = 20/$
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.
Leave a Reply