Projectile Motion - Quick Revision Notes (Lecture 05)

2-Dimensional Motion

  • Definition: The motion in which two of the three coordinates (x, y, z) change with time is called 2-Dimensional motion; it is planar motion.
  • In physics problems, projectile motion is a prime example of 2-D motion under gravity.

Projectile Motion

  • Definition: Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration due to gravity. The object is called a projectile and its path is the trajectory.
  • Acceleration is vertical, equal to \(g\) downward (≈ 9.8 m/s^2 or 10 m/s^2 in many worked examples). Horizontal motion has no acceleration (ignoring air resistance).

Terminologies

  • Velocity of Projection: The initial velocity with which a body is projected. Denoted by \(u\).
  • Launch Angle: The angle of the initial velocity with the horizontal, denoted by \(\theta\).
  • Horizontal Range: The horizontal distance travelled by the projectile during the entire motion. Denoted by \(R\).
  • Time of Flight: The total time the projectile spends in the air. Denoted by \(T\).
  • Maximum Height (h_max or H): The maximum vertical height reached by the projectile.

Key Equations (2-D kinematics)

  • Horizontal motion (constant velocity):
    • \(v_x = u \,\cos\theta\)
    • \(x = u \,\cos\theta \, t\)
  • Vertical motion (uniform acceleration -g):
    • \(v_y = u \,\sin\theta - g t\)
    • \(y = u \,\sin\theta \, t - \tfrac{1}{2} g t^2\)
  • Trajectory equation (eliminate time t):
    • From \(x = u \cos\theta \, t\), \(t = \frac{x}{u \cos\theta}\)
    • Substitute into y: \(y = x \tan\theta - \frac{g x^2}{2 u^2 \cos^2\theta}\)
  • Time of flight (to return to same vertical level):
    • \(T = \frac{2 u \sin\theta}{g}\)
  • Maximum height:
    • Occurs when vertical velocity becomes zero; time to reach max height: \(t_{top} = \frac{u \sin\theta}{g}\)
    • \(h_{max} = \frac{u^2 \sin^2\theta}{2 g}\)
  • Horizontal range (on level ground):
    • \(R = u \cos\theta \cdot T = \frac{u^2 \sin 2\theta}{g}\)
  • Relationship between range and maximum height:
    • \(\frac{R}{h_{max}} = \frac{u^2 \sin 2\theta / g}{u^2 \sin^2\theta / (2 g)} = 4 \cot\theta\)
    • Hence \(R = 4 \, h_{max} \, \cot\theta\)

Derivations and Key Results (selected concepts)

  • Horizontal component of velocity is constant during flight:
    • \(v_x = u \cos\theta\)
  • At the maximum height, vertical velocity is zero and the horizontal component remains unchanged:
    • \(vy = u \sin\theta - g t = 0 \Rightarrow t{top} = \frac{u \sin\theta}{g}\)
  • Example: Horizontal component at max height for any launch angle:
    • \(v_x = u \cos\theta\)
  • Example: If the range is 4√3 times the maximum height, the launch angle is:
    • From \(R = 4 h_{max} \cot\theta\
    • Given (R = 4\sqrt{3} h_{max}\), cot\theta = \sqrt{3} \Rightarrow \theta = 30^\circ\)
  • Example: Shortest range among multiple launch angles with the same speed occurs at the smallest value of sin(2θ):
    • Since \(R \propto \sin 2\theta\), the smallest sin 2θ among {15°, 30°, 45°, 60°} is at θ = 15°, so that body has the shortest range.
  • Example: Ball catching problem (kinematic matching):
    • If a runner runs at speed \(vc\) and a ball is projected with speed \(u\) at angle \(\theta\), the runner can catch the ball if the horizontal component of the ball equals the runner’s speed: \(u \cos\theta = vc\). Solve for θ given u and v_c.
  • Example: Projectile with velocity vector ((3\,\hat{i} + 10\,\hat{j})\) m/s:
    • Horizontal component: \(v_x = 3\,\text{m/s}\)
    • Initial vertical component: \(v_{y0} = 10\,\text{m/s}\)
    • Maximum height: \(h{max} = \frac{v{y0}^2}{2 g} = \frac{10^2}{2 \cdot 10} = 5\text{ m}\)
    • Time of flight: \(T = \frac{2 v_{y0}}{g} = \frac{20}{10} = 2\text{ s}\)
    • Range: \(R = v_x \cdot T = 3 \cdot 2 = 6\text{ m}\)
  • Angle-from-equality-of heights (two projectiles with initial angles 45° and 60° reach same height):
    • Equal heights imply \(u1^2 \sin^2 45^\circ = u2^2 \sin^2 60^\circ\)
    • With (\sin^2 45^\circ = 1/2\) and (\sin^2 60^\circ = 3/4\),
    • Therefore \(\frac{u1}{u2} = \sqrt{\frac{\sin^2 60^\circ}{\sin^2 45^\circ}} = \sqrt{\frac{3/4}{1/2}} = \sqrt{\frac{3}{2}} \approx 1.225\)
  • Speed at maximum height equals half of the initial speed implies a specific angle:
    • If the speed at the top is half of the initial speed, then the horizontal component must equal the half of the speed: \(v{top} = vx = u \cos\theta = \tfrac{u}{2} \Rightarrow \cos\theta = \tfrac{1}{2} \Rightarrow \theta = 60^\circ\)
  • Equation of trajectory (experimental forms):
    • General form: \(y(x) = x \tan\theta - \frac{g x^2}{2 u^2 \cos^2\theta}\)
    • Another equivalent form (eliminating t): \(x = u \cos\theta \, t, \quad y = u \sin\theta \, t - \tfrac{1}{2} g t^2\)
    • From these you can derive the standard parabola shape of the path.

Worked Examples (selected from the transcript)

  • Example: A bullet is fired with velocity (u) making angle (60^\circ) with the horizontal. The horizontal component of velocity at the maximum height is:
    • Answer: \(v_x = u \cos 60^\circ = \tfrac{u}{2}\)
  • Example: A bullet is fired with velocity (u = 10\,\text{m/s}) at (45^\circ). The range is:
    • \(R = \frac{u^2 \sin 90^\circ}{g} = \frac{100}{g}\)
    • With (g = 10\,\text{m/s}^2\), R = 10\,\text{m}.\n- Example: If (u = 10\,\text{m/s}) at (60^\circ), find (R) and (T) (g = 10):
    • (T = \frac{2 u \sin 60^\circ}{g} = \frac{2\cdot 10 \cdot (\sqrt{3}/2)}{10} = \sqrt{3}\) s
    • \(R = \frac{u^2 \sin 120^\circ}{g} = \frac{100 \cdot \sin 120^\circ}{10} = 10 \cdot \sin 60^\circ = 10 \cdot \tfrac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66\text{ m}\)
  • Example: A ball is thrown with speed 25 m/s at 60° and the fielder is 50 m away horizontally. The height at x = 50 m is:
    • Use trajectory formula: \(y = x \tan\theta - \frac{g x^2}{2 u^2 \cos^2\theta}\)
    • For (x = 50, \theta = 60^\circ, u = 25, g = 9.8): compute to get approximately (y \approx 8.2\text{ m}) (as shown in the transcript)\)
  • Bonus: Equal-Height condition for different projectiles (alternate method):
    • Use trajectory formulas to set heights equal and solve for the required speed ratio or angle ratio.

Additional Connections and Practical Relevance

  • Independence of horizontal and vertical motions (ignoring air resistance) is a core principle: horizontal motion is uniform, vertical motion is uniformly accelerated.
  • Real-world relevance: ballistics, sports (basketball, football, cricket), water jets, fireworks, and any scenario involving projectiles.
  • The trajectory is a parabola; real-world deviations occur due to air resistance, wind, and variation of gravity with altitude.

Summary of Core Formulas to Memorize

  • Horizontal component and trajectory:
    • \(v_x = u \cos\theta, \quad x = u \cos\theta \, t\)
    • \(v_y = u \sin\theta - g t, \quad y = u \sin\theta \, t - \tfrac{1}{2} g t^2\)
    • \(y(x) = x \tan\theta - \frac{g x^2}{2 u^2 \cos^2\theta}\)
  • Time of flight, height, range:
    • \(T = \frac{2 u \sin\theta}{g}\)
    • \(h_{max} = \frac{u^2 \sin^2\theta}{2 g}\)
    • \(R = \frac{u^2 \sin 2\theta}{g}\)
  • Relationship between R and h_max:
    • \(\frac{R}{h_{max}} = 4 \cot\theta\)
  • Key problem-solving steps:
    • Write x(t) and y(t) using initial components.
    • If needed, eliminate t to get trajectory y(x).
    • Use Y(T) = 0 for range, or Y top for h_max, etc.
    • For special scenarios, use R ∝ u^2 and the angle-dependent expressions to compare cases.

Notable References from the Transcript (Lecture 05, Physics Wallah)

  • Topic: Quick Last Lecture Revision - Projectile Motion; Subtopics: Projectile motion; Problems on projectile motion; 2-Dimensional motion; Parabolic trajectory.
  • Visuals: Parabolic path, components of motion, and coordinate points (A, B, C, D) illustrating the trajectory in the plane.
  • Practice DPP (Work, Power and Energy) follows the projectile notes in the batch.
  • Final reminders: Next lecture goals include subtopics and review questions.

Quick Practice Checks

  • If R = 4 cot θ × hmax, find θ when R/hmax = 4√3:
    • cot θ = √3 → θ = 30°.
  • If the speed doubles (u → 2u) with the same angle, how does R change?
    • Since R ∝ u^2, R doubles to 4 times the original: R2 = 4 R1.
  • For a projectile launched from ground with u = 19.6 m/s at θ = 30°, compute T and R (g = 9.8):
    • T = (\frac{2 u \sin\theta}{g} = \frac{2 \cdot 19.6 \cdot 0.5}{9.8} = 2) s
    • R = (\frac{u^2 \sin 2\theta}{g} = \frac{(19.6)^2 \sin 60^\circ}{9.8} \approx 33.1\text{ m}) (example values vary with rounding)

Connections to Previous Concepts

  • Builds on 2D motion and kinematics: decomposition into horizontal and vertical components, constant acceleration motion, and the use of trigonometry to relate angle, velocity, and range.
  • Links to energy concepts via work-energy ideas in the next module (Work, Power and Energy) and how projectile motion is a classic case study without work done by gravity in the horizontal direction.

End of Notes

The key formulas for projectile motion include for horizontal motion: the horizontal velocity v<em>x=ucos⁡θv<em>x = u \cos\theta and horizontal displacement x=ucos⁡θ tx = u \cos\theta \, t. For vertical motion: the vertical velocity v</em>y=usin⁡θ−gtv</em>y = u \sin\theta - g t and vertical displacement y=usin⁡θ t−12gt2y = u \sin\theta \, t - \tfrac{1}{2} g t^2. The trajectory equation is given by y=xtan⁡θ−gx22u2cos⁡2θy = x \tan\theta - \frac{g x^2}{2 u^2 \cos^2\theta}. Important parameters are calculated using: Time of Flight T=2usin⁡θgT = \frac{2 u \sin\theta}{g}, Maximum Height $$h_{max} =