Projectile Motion – Comprehensive Bullet-Point Study Notes

Overview of Kinematics Topics

• Chapter/lecture sequencing (as per syllabus)
  • 3.1 Linear Motion
  • 3.2 Motion With Constant Acceleration
  • 3.3 Free-Fall Motion
  • 3.4 Projectile Motion ⇦ current focus
  • 3.5 Worked Examples & Past-Year Paper Discussion
• Projectile Motion is the first full two-dimensional (2-D) application of earlier 1-D constant-acceleration theory.
  → All standard kinematic equations still apply independently in each perpendicular direction.

Physical Model of a Projectile

• Definition: Motion of any object launched into the air and subsequently influenced only by gravity (air resistance neglected).
• Typical sources: balls, bullets, arrows, rescue projectiles, etc.
• Coordinate choice:
  • $x$-axis → horizontal; acceleration $a_x = 0$
  • $y$-axis → vertical upward; acceleration $a_y = -g\;(g\approx 9.81\,\text{m\,s}^{-2})$
• Launch parameters:
  • Initial speed $v_0$
  • Launch angle (from horizontal) $\theta$
• Decomposition of initial velocity:
  • Horizontal component: $v<em>{0x}=v</em>0\cos\theta$
  • Vertical component: $v<em>{0y}=v</em>0\sin\theta$
• Fundamental idea: Treat the 2-D path as two simultaneous 1-D motions—uniform motion horizontally, uniformly accelerated motion vertically.

Horizontal-Direction Kinematics (Uniform Motion)

• Displacement: $x = v<em>{0x} t = (v</em>0\cos\theta)\,t$
  → Time of flight can be isolated: $t = \dfrac{x}{v_0\cos\theta}$
• Velocity: $v<em>x(t)=v</em>{0x}=v<em>0\cos\theta$ (constant because $a</em>x=0$)
• Acceleration: $a_x = 0$ (no horizontal force in the ideal model).

Vertical-Direction Kinematics (Uniform Acceleration)

• Displacement: $y = v<em>{0y}t - \tfrac12 g t^2 = (v</em>0\sin\theta)t - \tfrac12 g t^2$
• Velocity: $v<em>y(t)=v</em>{0y}-gt = v_0\sin\theta - g t$
• Acceleration: $a_y = -g$ (constant, acts downward).

Trajectory Equation (Eliminating Time)

1. Express $t$ from horizontal motion: $t = \dfrac{x}{v_0\cos\theta}$
2. Substitute into vertical displacement:
   $y = (v<em>0\sin\theta)\left(\dfrac{x}{v</em>0\cos\theta}\right) - \tfrac12 g \left( \dfrac{x}{v_0\cos\theta} \right)^2$
3. Simplify:
   $y = x\tan\theta - \frac{g x^2}{2v_0^2 \cos^2\theta}$

• Significance: quadratic in $x$, showing that an ideal projectile travels in a parabolic path.

Key Performance Quantities

1. Range, $R$ (Horizontal distance when the projectile returns to launch height)

• Condition: $y = 0$ when $x = R$.
• Solve trajectory equation with $y=0$:
  $R\tan\theta - \frac{g R^2}{2v<em>0^2\cos^2\theta}=0$ → $R\left[\tan\theta - \frac{g R}{2v</em>0^2\cos^2\theta}\right]=0$
  → Non-trivial solution:
  $R = \frac{v_0^2\sin2\theta}{g}$
• Practical note: Maximum range for a given $v_0$ occurs at $\theta=45^\circ$ (because $\sin2\theta$ peaks at 1).

2. Maximum Height, $H$

Approach A (use vertical velocity at apex):

• At highest point $v<em>y = 0$. Use $v</em>y^2 = v<em>{0y}^2 - 2gH$: $0 = (v</em>0\sin\theta)^2 - 2gH \;\Rightarrow\; H = \frac{v_0^2\sin^2\theta}{2g}$

Approach B (via trajectory equation at $x=R/2$):

• Symmetry: apex occurs halfway horizontally so $x=R/2$ → substitute back; yields identical result above.

3. Time of Flight, $T$ (to return to launch height)

• Use vertical motion: $y=0$ again, but $t\ne0$:
  $0 = v<em>{0y}T - \tfrac12 g T^2 \;\Rightarrow\; T( v</em>{0y} - \tfrac12 g T)=0$
  → $T = \frac{2v_0\sin\theta}{g}$

Interconnections & Conceptual Remarks

• Equations $R,\;H,\;T$ each involve $v_0$, $\theta$, and $g$; knowing any two usually determines the third.
• Real-world deviations (air drag, wind, spin) introduce horizontal acceleration and/or variable vertical acceleration; model must then be extended.
• Ethically and practically, projectile calculations underlie sports, rescue operations, ballistics, and even space-mission trajectory planning.

Worked-Example Synopses (Slide References)

Example 1 – Golf Ball ($v_0 = 30\,\text{m\,s}^{-1}, \; \theta = 30^\circ$)

• Tasks: (a) components $v<em>{0x}, v</em>{0y}$; (b) velocity at $y=0.5\,\text{m}$; (c) $H$; (d) $R$; (e) $T$.
• Recall formulas:
  $v<em>{0x}=30\cos30^\circ\approx25.98$ m/s, $v</em>{0y}=30\sin30^\circ=15.0$ m/s, etc.

Example 2 – Rifle Bullet ($v_0 = 250\,\text{m\,s}^{-1}, \; \theta = 37^\circ$)

• Find relative peak height (H), range (R), and time aloft (T).
  • $H = \dfrac{(250)^2\sin^2 37^\circ}{2g}$, etc.

Example 3 – Quarterback/Receiver Problem

• Given: $v_0=20\,\text{m\,s}^{-1}, \theta=30^\circ,\; \text{initial separation}=20\,\text{m}$.
• Objective: receiver’s running speed & direction so he meets ball at launch height.
  • First compute $T$, then required horizontal running speed $v_r = \dfrac{\text{additional horizontal distance}}{T}$.

Example 4 – Mountain-Climber Rescue (Ledge 30 m high, 50 m away, $\theta=55^\circ$)

• Unknown: $v_0$ such that projectile lands precisely on ledge.
• Use simultaneous equations: $y=30\,\text{m},\;x=50\,\text{m}$.

Example 5 – Horizontal Launch from Bridge

• Pure horizontal initial velocity $v_0=7.0\,\text{m\,s}^{-1},\; h=30\,\text{m}$.
• Find fall time $T =\sqrt{\dfrac{2h}{g}}$, impact velocity vector, and horizontal range $x=v_0T$.

Example 6 – Cannon (40°, 100 m s⁻¹)

• Components, peak altitude, and time to reach $y=100\,\text{m}$ on ascent are required.
• Use vertical motion equation solved for $t$ when $y=100\,\text{m}$ with positive root < apex time.

Example 7 – Two Balls, Equal Apex Height

• Straight-up ball returns in 3 s ⇒ apex reached at $1.5\,\text{s}$, yielding $H = \tfrac12 g (1.5)^2$.
• Second ball launched at 30° must have $v<em>{0y} = \sqrt{2gH}$; relate to $v</em>0$ via $\sin30^\circ$ to obtain required speed.

Example 8 – Building (35 m) Horizontal Throw

• Impacts ground 80 m horizontally away. Solve $T=\sqrt{2h/g}$, find $v_{0x}=x/T$, compute impact velocity.

Example 9 – Stone Thrown From 45-m Building ($\theta=30^\circ, v_0=20\,\text{m\,s}^{-1}$)

• Multi-part: flight time until it hits ground, final speed, horizontal landing distance $x$, and velocity at $y=30\,\text{m}$.
• Altered reference level: initial $y=+45\,\text{m}$ relative to ground.

Example 10 – Baseball Home-Run (clears 21 m wall 130 m away, $\theta=35^\circ$)

• Unknown $v_0$, time to wall, and velocity components at wall height.
• Use simultaneous $x,y$ equations with offset: initial launch height $1\,\text{m}$.

Past-Year Paper Questions (Notation “B1 (0405)” etc.)

• B1 (a) asks for $v<em>{0x}, v</em>{0y}$ in terms of $v$ and $\theta$.
• B1 (b) requests derivation of trajectory equation $y(x)$; worth $3.5$ marks.
• B1 (c) calculates height at given $x$, range, and maximum height for $v_0=50\,\text{m\,s}^{-1},\;\theta=30^\circ$.
• B2 (friction question) included for broader syllabus continuity:
  • Four properties of frictional force.
  • Derivation of $\mu_s = \tan\theta$ for impending slip on an incline.

Consolidated Formula Sheet (quick reference)

• Components:
  $\boxed{v<em>{0x}=v</em>0\cos\theta}\qquad\boxed{v<em>{0y}=v</em>0\sin\theta}$
• Horizontal motion:
  $x=v<em>{0x} t\quad( a</em>x=0 )$
• Vertical motion:
  $y=v<em>{0y} t - \tfrac12 g t^2$$v</em>y=v<em>{0y}-g t$$v</em>y^2 = v_{0y}^2 - 2g y$
• Time of flight (level ground): $\boxed{T=\dfrac{2v_0\sin\theta}{g}}$
• Range (level ground): $\boxed{R=\dfrac{v_0^2\sin2\theta}{g}}$
• Maximum height: $\boxed{H=\dfrac{v_0^2\sin^2\theta}{2g}}$
• Trajectory: $\boxed{y = x\tan\theta - \dfrac{g x^2}{2v_0^2\cos^2\theta}}$

Practical & Ethical Connections

• Sports analytics: optimizing launch angles for javelin, golf, basketball.
• Search-and-rescue: delivering payloads (food, rope) via parabolic arc to inaccessible locations (Example 4).
• Ballistics & safety: accurate gunfire or protective barriers rely on understanding of range & trajectory.
• Space missions: upper stages initially behave like high-speed projectiles before orbital insertion.
• Ethical obligation: civil engineers and weapon designers must consider safety, unintended consequences, and legal regulations when applying projectile theory.

Study Tips

• Always draw a diagram indicating $x$ and $y$ axes, initial components, and any reference heights.
• Check units; keep $g$ consistent (SI: $9.81\,\text{m\,s}^{-2}$ or approximate $9.8$ as allowed by exam rubric).
• When solving unfamiliar configurations (launch/landing heights differ), keep vertical displacement sign-convention consistent.
• For maximum-range or max-height proofs, recognize trigonometric identities $\sin2\theta = 2\sin\theta\cos\theta$.
• Memorize core formulas; derive others quickly from first principles to minimize rote errors.