Problem Statement
Solve the kinematics problem: A ball is thrown horizontally at 10 m/s from a point 45 m above the ground on a vertical cliff. Find the horizontal range. ($g = 10$ m/s²) $t = \sqrt{2h/g}$; Range $= v_x \times t$ Step 1: $t = \sqrt{2(45)/10} = \sqrt{9} = 3$ s. Step 2: Range $= 10 \times 3 = 30$ m. $$\boxed{R = 30\text{ m}}$$
Given Information
- $\boxed{R = 30\text{ m}$
Physical Concepts & Formulas
Friction is a contact force opposing relative motion (kinetic friction) or impending motion (static friction). On an inclined plane, the weight component along the slope is $mg\sin\theta$ and the normal force is $N = mg\cos\theta$, giving maximum static friction $f_{s,\max} = \mu_s mg\cos\theta$. The condition for sliding is $\tan\theta > \mu_s$.
- $f = \mu N$ — kinetic friction force
- $N = mg\cos\theta$ — normal force on incline
- $mg\sin\theta – \mu mg\cos\theta = ma$ — Newton’s 2nd law along incline
- $\tan\theta_c = \mu_s$ — critical angle for sliding
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
$$\boxed{R = 30\text{ m}}$$
$$R = \frac{u^2\sin 2\theta}{g} = \frac{400\times\sin 60°}{9.8} = \frac{400\times0.866}{9.8} = \frac{346.4}{9.8} \approx 35.3\,\text{m}$$
$$H = \frac{u^2\sin^2\theta}{2g} = \frac{400\times0.25}{19.6} = \frac{100}{19.6} \approx 5.1\,\text{m}$$
Answer
$$\boxed{R = 30\text{ m}}$$
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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