1.5

Overview of Projectile Motion

• Objects launched into the air and subjected to gravity are called projectiles.

• The path of a projectile is a parabola (an arch) due to constant downward acceleration from gravity.

• The launch angle affects how far (range) and how high the projectile will travel; trig identities are used to resolve motion into components.

• Under idealized conditions (no air resistance), analysis treats horizontal and vertical motions separately but with the same flight time.

• Key formulas (under standard orientation with upward as +y and to the right as +x):

  • Initial velocity components:

  • $v<em>{0x} = v</em>0 \cos\theta$

  • $v<em>{0y} = v</em>0 \sin\theta$

  • Horizontal motion (x-direction): constant velocity, no horizontal acceleration

  • $x(t) = x<em>0 + v</em>{0x} t$

  • $v<em>x(t) = v</em>{0x}$

  • $a_x = 0$

  • Vertical motion (y-direction): uniform acceleration due to gravity, downward is negative

  • $y(t) = y<em>0 + v</em>{0y} t - \tfrac{1}{2} g t^2$

  • $v<em>y(t) = v</em>{0y} - g t$

  • $a_y = -g\quad (g \approx 9.8\ \text{m/s}^2\text{ or }10\ \text{m/s}^2\text{ in many problems})$

• Parabolic trajectory (path equation) can be found by eliminating time t between x and y:

  • $y = x \tan\theta - \frac{g x^2}{2 v_0^2 \cos^2\theta}$

  • This shows that the trajectory is a parabola in the x–y plane.

• Special case: Horizontally launched projectiles

  • Launch angle: (\theta = 0) → vertical component is zero: (v_{0y} = 0)

  • Horizontal velocity is constant: $v<em>x(t) = v</em>{0x}$

  • Vertical motion is independent and governed by gravity: $y(t) = y_0 - \tfrac{1}{2} g t^2$ (using upward as positive; ground reached when y = 0)

Two-Dimensional Motion and Components

• Projectiles move in two dimensions: x (horizontal) and y (vertical).

• Each dimension has its own displacement and velocity:

  • Horizontal component represents how far it flies; magnitude remains constant if there is no air resistance.

  • Vertical component represents how high it flies; magnitude changes due to gravity and direction changes as it moves upward and then downward.

• Time is the same for both x and y components (flight time).

Horizontal vs Vertical Components

• Horizontal component:

  • Velocity is constant: $v<em>x = v</em>{0x} = \text{constant}$

  • Acceleration is zero: $a_x = 0$

  • Gravity does not change horizontal velocity.

• Vertical component:

  • Velocity changes due to gravity: $v<em>y(t) = v</em>{0y} - g t$

  • Magnitude and direction change during flight:

  • While moving up, magnitude decreases and direction is upward.

  • While moving down, magnitude increases and direction is downward.

• Combining components yields the trajectory (parabolic path).

• Components at a glance:

  • Horizontal: Magnitude = Constant; Direction = Constant

  • Vertical: Magnitude = Changes; Direction = Changes

Trajectory and Equations of Motion

• The trajectory is parabolic due to the constant downward acceleration.

• General 2D motion equations (assuming initial position at origin for simplicity):

  • $x(t) = x<em>0 + v</em>{0x} t$

  • $y(t) = y<em>0 + v</em>{0y} t - \tfrac{1}{2} g t^2$

• If starting from the origin with x0 = 0, y0 = 0 and launch angle (\theta) and speed (v_0):

  • $v<em>{0x} = v</em>0 \cos\theta,\quad v<em>{0y} = v</em>0 \sin\theta$

  • Path equation: $y = x \tan\theta - \frac{g x^2}{2 v_0^2 \cos^2\theta}$

• The time of flight and range depend on initial conditions (height, angle, speed).

Horizontally Launched Projectiles

• Characteristics:

  • No upward trajectory and no initial vertical velocity: (v_{0y} = 0).

  • Horizontal velocity is constant: $v<em>x = v</em>{0x} = \text{constant}$

  • Vertical motion is a free fall under gravity from height (y_0):

  • $y(t) = y_0 - \tfrac{1}{2} g t^2$ (upward positive convention)

• Analysis requires two equations (one for x, one for y) and the same time variable t.

  • Horizontal: $x(t) = x<em>0 + v</em>{0x} t$

  • Vertical: $y(t) = y<em>0 - \tfrac{1}{2} g t^2$ (if (v{0y}=0))

• Practical approach:

  • Solve vertical motion to find time (t) when the projectile hits the ground ((y = 0) or the ground level).

  • Use that time in the horizontal equation to find the horizontal distance traveled.

Sample Problem: Cannon on a Cliff (Horizontal Launch)

• Given: A cannon on a cliff height (H = 60\ \text{m}) above the ground.

• Initial horizontal velocity: (v_{0x} = 95\ \text{m/s}).

• Gravity: (g = 10\ \text{m/s}^2).

• Task: Find how far from the base of the cliff the projectile lands.

• Solution steps:

  • Step 1: Vertical motion (no initial vertical velocity):

  • Use ground level y = 0 with initial height y_0 = H:

  • $0 = H - \tfrac{1}{2} g t^2 = 60 - \tfrac{1}{2} (10) t^2$

  • Solve for (t): (t^2 = \dfrac{2H}{g} = \dfrac{2\cdot 60}{10} = 12) ⇒ (t = \sqrt{12} \approx 3.46\ \text{s}).

  • Step 2: Horizontal distance (range):

  • $x = v_{0x} t = 95 \times 3.46 \approx 329\ \text{m}$

  • Step 3: Interpret and record results; draw labeled diagram; fill givens table as practice.

• Notes on this problem:

  • The vertical motion determines time of flight; horizontal motion determines distance.

  • Time found from vertical equation is independent of horizontal velocity (as long as air resistance is neglected).

Time of Flight and Horizontal Distance (Derived Equations)

• General time of flight for a horizontal launch from height (H) with zero initial vertical velocity:

  • $t = \sqrt{\frac{2H}{g}}$

• General horizontal distance during that flight:

  • $D = v<em>{0x} t = v</em>{0x} \sqrt{\frac{2H}{g}}$

• For angled launches (nonzero (v_{0y})), the time to hit the ground is obtained from:

  • $y<em>0 + v</em>{0y} t - \tfrac{1}{2} g t^2 = 0$

  • Solve for the positive root of this quadratic to obtain the flight time, then use in the horizontal equation to get the range.

Baby’s First Derived Equation (Horizontal Throw from Height)

• Scenario: A rock is thrown horizontally with speed (v) from the top of a cliff of height (H).

• Part A: Establish coordinate system and directions (x and y origins and positive directions).

• Part B: Identify the time equation (time to reach water/ground):

  • From vertical motion: $t = \sqrt{\frac{2H}{g}}$

• Part C: Rearrange to express time in terms of given quantities: the same formula as Part B is the final form for time.

• Part D: Identify the horizontal distance to impact: $D = v \cdot t = v \sqrt{\frac{2H}{g}}$

• Takeaway: Horizontal motion decouples from vertical motion; time depends only on height and gravity for horizontal launch.

Connections to Foundational Principles and Real-World Relevance

• Independence of perpendicular motions: the x- and y- motions evolve simultaneously but independently, linked only by the shared time variable.

• Gravity acts only in the vertical direction (in the idealized model) and is constant, producing a parabola for the trajectory.

• Real-world relevance:

  • Sports (basketball shots, football throws, soccer kicks)

  • Ballistics and engineering (projectile trajectories in design considerations, safety analyses)

  • Computer simulations and video game physics use these same equations for believable motion.

Quick Reference: Key Formulas to Memorize

• Components of initial velocity:

  • $v<em>{0x} = v</em>0 \cos\theta$

  • $v<em>{0y} = v</em>0 \sin\theta$

• Position as a function of time:

  • Horizontal: $x(t) = x<em>0 + v</em>{0x} t$

  • Vertical: $y(t) = y<em>0 + v</em>{0y} t - \tfrac{1}{2} g t^2$

• Velocity components as a function of time:

  • $v<em>x(t) = v</em>{0x}$

  • $v<em>y(t) = v</em>{0y} - g t$

• Parabolic trajectory equation (eliminate t):

  • $y = x \tan\theta - \frac{g x^2}{2 v_0^2 \cos^2\theta}$

• Time of flight for horizontal launch from height H (v_{0y} = 0):

  • $t = \sqrt{\frac{2H}{g}}$

• Horizontal range for horizontal launch from height H:

  • $D = v<em>{0x} t = v</em>{0x} \sqrt{\frac{2H}{g}}$

• Sample problem result (H = 60 m, v0x = 95 m/s, g = 10 m/s^2):

  • Time: $t = \sqrt{\frac{2 \cdot 60}{10}} = \sqrt{12} \approx 3.46\ \text{s}$

  • Range: $D = 95 \times 3.46 \approx 329\ \text{m}$

• Important conceptual ideas to remember:

  • The horizontal component of velocity remains constant in the absence of air resistance.

  • The vertical component is governed by gravity and changes over time.

  • The overall motion is the combination of these two independent components, producing a parabola.

• Common problem-solving approach:

  • Step 1: Decompose the initial velocity into components.

  • Step 2: Write the x- and y-motion equations.

  • Step 3: Use the vertical motion to find the time of flight (often by setting y to the ground level).

  • Step 4: Substitute the time into the horizontal motion to find the range.

  • Step 5: If height or angle changes, adjust the equations accordingly and solve for the desired quantity.