Problem Statement
Solve the kinematics problem: A ball is projected with speed 20 m/s at 30° above horizontal. Find the horizontal range. ($g = 10$ m/s²) $R = u^2\sin 2\theta / g$ Step 1: $R = \dfrac{(20)^2 \sin 60°}{10} = \dfrac{400 \times (\sqrt{3}/2)}{10} = 40\sqrt{3}/2 = 20\sqrt{3} \approx 34.6$ m. $$\boxed{R = 20\sqrt{3} \approx 34.6\text{ m
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Projectile motion decomposes into independent horizontal and vertical components. Horizontal: constant velocity (no air resistance). Vertical: constant downward acceleration $g$. The trajectory is a parabola. Maximum range occurs at $45°$ launch angle; max height at $90°$.
- $x = v_0\cos\theta \cdot t$, $y = v_0\sin\theta \cdot t – \frac{1}{2}gt^2$
- $R = v_0^2\sin 2\theta/g$ — horizontal range
- $H = v_0^2\sin^2\theta/(2g)$ — maximum height
- $T = 2v_0\sin\theta/g$ — total flight time
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\boxed{R = 20\sqrt{3} \approx 34.6\text{ m
Given Information
- Initial velocity $u$ (or $v_0$)
- Acceleration $a$ (constant unless stated otherwise)
- Time $t$ or distance $s$ as given
Physical Concepts & Formulas
Kinematics describes motion without reference to its cause. For constant acceleration, the four SUVAT equations are sufficient to solve any problem. They follow directly from the definitions of velocity ($v = ds/dt$) and acceleration ($a = dv/dt$). For 2D problems (projectile motion), the horizontal and vertical motions are independent — horizontal: constant velocity; vertical: constant acceleration $g$ downward. Relative motion problems require defining a reference frame explicitly and using vector subtraction.
- $v = u + at$
- $s = ut + \tfrac{1}{2}at^2$
- $v^2 = u^2 + 2as$
- $s = \tfrac{1}{2}(u+v)t$
- Range of projectile: $R = \dfrac{u^2\sin 2\theta}{g}$
- Max height: $H = \dfrac{u^2\sin^2\theta}{2g}$
Step-by-Step Solution
Step 1 — List knowns and unknown: $u$, $v$, $a$, $s$, $t$ — identify which three are known.
Step 2 — Choose the SUVAT equation that contains the unknown and all three known quantities.
Step 3 — Substitute and solve algebraically.
Step 4 — For 2D: Decompose $\vec{u}$ into $u_x = u\cos\theta$, $u_y = u\sin\theta$. Solve $x$ and $y$ separately.
Worked Calculation
Substituting all values with units:
Projectile at $u = 20\,\text{m/s}$, $\theta = 30°$:
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Answer
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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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