Projectile Motion Comprehensive Study Notes

Definition of Projectile Motion

A projectile is any object that, once launched, moves under the influence of gravity alone. In the idealised model used in class the following hold:

The only external force acting after launch is the weight $W=mg$ , directed vertically downward.
Air resistance is neglected, so no horizontal (drag) or additional vertical forces appear in the equations.
The motion is therefore two–dimensional and can be decomposed into independent horizontal and vertical components.

Assumptions (Negligible Air Resistance)

Gravity is uniform: the acceleration due to gravity $g$ is constant (≈ $9.81\,\text{m s}^{-2}$ near Earth’s surface).
The launch and landing points are usually measured from a common reference line (often the ground level). If the landing point differs in height, an extra vertical displacement term is included.
No spin, no lift, no Magnus effect: the projectile neither produces lift nor changes its shape.

Fundamental Equations of Motion (in 2-D)

The standard constant-acceleration kinematic relations apply separately to the x- and y-axes:

Horizontal direction (no acceleration):
$x(t)=v_{0x}\,t$
Vertical direction (constant downward acceleration $-g$ ):
$y(t)=v{0y}\,t-\frac12 g t^2$ $vy(t)=v{0y}-g t$ Here $v{0x}=v0\cos\theta$ and $v{0y}=v0\sin\theta$ , where $v0$ is the launch speed and $\theta$ the launch angle above the horizontal.

Choosing a Coordinate System & Sign Conventions

In class the downward direction was taken as negative, hence the vertical acceleration is negative $(-g)$ . The student’s question “Why is she using the negative?” referred exactly to this convention.
You may equally choose upward positive, provided you remain consistent throughout the calculation.

Decomposing the Initial Velocity

The class repeatedly emphasised resolving the initial velocity vector:
$v{0x}=v0\cos\theta, \qquad v{0y}=v0\sin\theta$
Key reason: the x- and y-motions are independent once resolved.

Time of Flight (Same Launch & Landing Height)

Derivation: Set final vertical displacement $y(tf)=0$ and solve. $0=v{0y}tf-\frac12 g tf^2 \implies tf=\frac{2v{0y}}{g}=\frac{2v_0\sin\theta}{g}$

Maximum Height

Occurs when instantaneous vertical velocity is zero $vy=0$ : $0=v{0y}-gt{H} \implies t{H}=\frac{v{0y}}{g}$ Substitute $tH$ back into $y(t)$ :
$H=\frac{v{0y}^2}{2g}=\frac{v0^2\sin^2\theta}{2g}$

Horizontal Range (Same Launch & Landing Height)

Plug $tf$ into $x(t)$ : $R=v{0x}tf=v0\cos\theta\left(\frac{2v0\sin\theta}{g}\right)=\frac{v0^2\sin2\theta}{g}$
Note the appearance of $\sin2\theta$ : maximum range for a given $v_0$ occurs at $\theta=45\,^{\circ}$ .

Trajectory Equation (Complete Path)

Eliminate time $t$ between $x(t)$ and $y(t)$ :
$t=\frac{x}{v{0x}}$ leads to $y=x\tan\theta-\frac{g x^2}{2 v0^2\cos^2\theta}$
This is a quadratic (parabolic) curve opening downward, explaining why every ideal projectile follows a parabolic arc.

Worked Classroom Example (Values from the Discussion)

The transcript showed an in-class calculation; the blackboard contained partial numerical results:

Angle read off: $\theta≈25^{\circ}$ (student mentions “twenty over twenty-five”).
Time of flight: $t_f≈0.927\,\text{s}$ (one student reported $0.92\,\text{s}$ ).
Maximum height: $H≈1.30\,\text{m}$ (quoted “1.3”).
Horizontal range: $R≈12.86\,\text{m}$ (value “twelve point eight six”).
Another vertical displacement printed on the board: $1.81\,\text{m}$ (likely an intermediate check at $t=0.5\,\text{s}$ ).

Although not all intermediate numbers were read clearly, the steps likely followed the template:

Resolve $v0$ into $v{0x}, v_{0y}$ using the given angle.
Compute $tf$ via $tf=\frac{2v_{0y}}{g}$ .
Compute $H$ and $R$ with the formulas above.
Insert negative sign correctly in $y(t)$ because acceleration is downward.

Common Mistakes & Clarifications

Forgetting the sign of $g$ : leads to wrong height or time values. The student’s “Why is she using the negative?” calls attention to this.
Mixing degree and radian mode on calculators (teacher reminded the class “I’m supposed to be using degrees, right?”).
Using the wrong angle (complement instead of the actual launch angle).
Omitting units: always state s, m, or $\text{m s}^{-1}$ .
Plugging horizontal velocity into a vertical formula and vice-versa.

Connections to Prior Learning & Real-World Applications

Builds directly on 1-D constant-acceleration kinematics. Concepts like “independent components” mirror vector addition from previous lectures.
Practical: ballistics, sports (basketball arc, golf drive), firefighting hose angles, and any engineering context where trajectories must avoid obstacles.
The independence principle also underlies modern computer-graphics engines and animation.

Ethical, Philosophical & Practical Considerations

Historically, projectile work was driven by military needs; teaching must balance appreciation for the physics with awareness of potential misuse.
Understanding parabolic motion encourages safe design of recreational equipment and public infrastructure (e.g., water fountains, skate-park ramps).
Emphasises the philosophical idea that simple, universal laws (Newton’s) can predict apparently complex motion.

Numerical & Symbolic Summary of Key Formulas

$x(t)=v_0\cos\theta\;t$
$y(t)=v_0\sin\theta\;t-\frac12 g t^2$
$tf=\frac{2v0\sin\theta}{g}$
$H=\frac{v_0^2\sin^2\theta}{2g}$
$R=\frac{v_0^2\sin2\theta}{g}$
$y=x\tan\theta-\frac{g x^2}{2 v_0^2\cos^2\theta}$
Memorise these, but also practise deriving them to reinforce understanding.