Problem Statement
Solve the kinematics problem: A projectile is launched at speed $u$ and angle $\theta$. At what time does the speed equal $u\cos\theta$? At the highest point, $v_y = 0$ and $v = v_x = u\cos\theta$ Step 1: The speed equals $u\cos\theta$ when $v_y = 0$, i.e., at the highest point. Step 2: Time to reach top: $v_y = u\sin\theta – gt
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
Full substitution shown in the steps above.
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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