Problem Statement
Solve the kinematics problem: Find the linear speed of a point on the Earth’s equator due to Earth’s rotation. ($R_E = 6400$ km) All quantities, constants, and constraints stated in the problem above Physical constants used as needed (see Concepts section) This problem draws on fundamental physical principles. The key is to iden
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 $u = 20\,\text{m/s}$, $\theta = 30°$:
$$R = \frac{u^2\sin 2\theta}{g} = \frac{400\times\sin 60°}{9.8} = \frac{400\times0.866}{9.8} = \frac{346.4}{9.8} \approx 35.3\,\text{m}$$
$$H = \frac{u^2\sin^2\theta}{2g} = \frac{400\times0.25}{19.6} = \frac{100}{19.6} \approx 5.1\,\text{m}$$
Answer
$$\boxed{R = \dfrac{u^2\sin 2\theta}{g},\quad H = \dfrac{u^2\sin^2\theta}{2g}}$$
Physical Interpretation
Maximum range occurs at $\theta = 45°$ where $\sin 90°=1$. The complementary angles $30°$ and $60°$ give the same range — a 20 m/s ball at either angle reaches ~35 m. Athletes intuitively throw at 45° for distance. The horizontal range is quadratic in $u$, so doubling the speed quadruples the range.
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