Notes: Vectors and Projectile Motion

Vector Basics

Magnitude is a scalar property of a vector: it is the size or amount without direction.
A scalar has only magnitude; a vector has both magnitude and direction.
If two vectors have equal magnitudes and opposite directions, their sum can be zero.
- Example intuition: when two equal pushes/pulls cancel each other out, the resulting vector is zero.
Instantaneous speed vs instantaneous velocity:
- Instantaneous speed: how fast something is moving at a given moment (magnitude only).
- Instantaneous velocity: speed with a specified direction (a vector).
Notation for a vector: different ways of writing the same quantity, e.g.
- \textit{-30 m/s west} vs +30 m/s east represent opposite directions; the vector has magnitude 30 and direction.
Adding a vector to a scalar is not defined: you cannot add a quantity with direction to a quantity with no direction.

Basic vector example (A and B)

Given:
- Vector A has length 3.00 units along the positive x-axis: $\boldsymbol{A} = (3, 0)$
- Vector B has length 4.00 units along the negative y-axis: $\boldsymbol{B} = (0,-4)$
Compute combinations:
- a) \boldsymbol{A} + \boldsymbol{B} = (3, -4)
- Magnitude: $|\boldsymbol{A}+\boldsymbol{B}| = \sqrt{3^2 + (-4)^2} = 5.0$
- Direction: $ heta = an^{-1}igg(\frac{-4}{3}\bigg) = -53.13^{\circ} $ relative to the +x axis (i.e., 53.13° below +x)
- b) \boldsymbol{A} - \boldsymbol{B} = (3, 4)
- Magnitude: $|\boldsymbol{A}-\boldsymbol{B}| = \sqrt{3^2 + 4^2} = 5.0$
- Direction: $ \theta = \tan^{-1}(4/3) = 53.13^{\circ} $ above the +x axis
- c) \boldsymbol{A} + 2\boldsymbol{B} = (3, -8)
- Magnitude: $|\boldsymbol{A}+2\boldsymbol{B}| = \sqrt{3^2 + (-8)^2} = \sqrt{73} \approx 8.54$
- Direction: $ \theta = \tan^{-1}(-8/3) = -69.44^{\circ} $ (69.44° below +x)
- d) \boldsymbol{B} - \boldsymbol{A} = (-3, -4)
- Magnitude: $|\boldsymbol{B}-\boldsymbol{A}| = 5.0$
- Direction: in the third quadrant; $ \theta = \tan^{-1}((-4)/(-3)) = 53.13^{\circ} $ below the negative x-axis, or 233.13° from the +x axis
Quick note on Pythagorean relation: for vectors with components $(a,b)$, the magnitude is $\sqrt{a^2 + b^2}.$

Vector Arithmetic (specific example with A=(3,0), B=(0,-4))

A = (3,0) along +x; B = (0,-4) along -y.
Results:
- (a) A + B = (3,-4): magnitude 5.0; direction 53.13° below +x.
- (b) A - B = (3,4): magnitude 5.0; direction 53.13° above +x.
- (c) A + 2B = (3,-8): magnitude ≈ 8.54; direction 69.44° below +x.
- (d) B - A = (-3,-4): magnitude 5.0; direction 53.13° below the -x axis (or 233.13° from +x).
The angle calculations often use \tan^{-1}; for example, arctan(4/3) ≈ 53.13° and arctan(8/3) ≈ 69.44°.

Projectile Motion: Launch and Velocity Components

Define:
- Va: muzzle/launch speed (magnitude of initial velocity)
- θ: launch angle with respect to +x axis
- Vx: horizontal component, $Vx = Va \cos\theta$
- Vy: vertical component, $Vy = Va \sin\theta$
- Gravity g ≈ 9.81 m/s^2 downward
At launch: velocity components are (Vx, Vy).
At the top (highest point) of the trajectory: Vy = 0.
Acceleration is constant downward: $a_y = -g.$
Horizontal motion is uniform: $Vx(t) = Vx = V_a \cos\theta.$
Vertical motion is uniformly accelerated: $Vy(t) = Va \sin\theta - g t.$
Key times:
- Time to reach the top: $t{top} = \frac{Va \sin\theta}{g}$
- Time to rise to apex equals time to fall back to launch height (symmetry) when landing height = launch height.
- Total time of flight (landing at same height): $T = \frac{2 V_a \sin\theta}{g}$
Projectile range on level ground: $R = \frac{V_a^2 \sin(2\theta)}{g}.$ (references: standard kinematics for projectile motion)
Special cases and observations from the notes:
- With no air resistance, the vertical and horizontal motions are decoupled: horizontal range depends on speed and angle via the formula above.
- For angles θ and 90°−θ, the ranges are equal: $R(\theta) = R(90^{\circ}-\theta)$ because sin(2θ) = sin(180°−2θ).
- Maximum range on level ground (with fixed Va) occurs at θ = 45°.
- If the speed is the same but angles are 40° and 50° (complementary to 90°), they yield the same range in the absence of air resistance.

Time and Distance Narratives in Projectile Motion

Vertical velocity magnitude changes as the projectile moves:
- From launch to the apex: Vy decreases from its initial value to 0 (if Vy0 > 0). If launching upward, Vy is positive early and decreases to zero.
- From apex to the ground: Vy becomes negative and increases in magnitude as the projectile falls.
Horizontal velocity magnitude remains constant throughout the flight (no air resistance): $|Vx| = Va \cos\theta.$
Vertical acceleration magnitude is constant: $|a_y| = g.$ Direction is downward.
Time to reach the top is half of the time to land for symmetric level-ground launches:
- Time to top: $t{top} = \dfrac{Va \sin\theta}{g}$
- Time to land (from launch to ground): T = \dfrac{2 Va \sin\theta}{g} = 2 t{top}.$n

Ballistic Scenarios and Intuition (No Air Resistance)

Two balls dropped from the same height, one second apart:
- After the second is released, the distance between them remains constant (because both experience the same acceleration).
A ball launched horizontally vs a ball dropped vertically from the same height:
- Both hit the ground after the same time (vertical motion is identical if released from same height with Vy0 = 0).
- The horizontal-velocity-carrying ball has a larger impact speed because it also has a horizontal component in addition to the vertical component at impact.
A luggage/package dropped from a moving airplane:
- Without air resistance, the package retains the plane’s forward velocity and lands ahead of the point directly beneath the drop line; it does not fall straight down vertically under the initial ground point.
- In practice, air resistance and wind alter landing location, but the initial horizontal velocity equals the plane’s speed at release.

Vector Magnitude, Direction, and Equal Components

A vector has equal horizontal and vertical components when its angle from the x-axis is 45°, i.e., if the vector makes a 45° angle with the x-axis, then |Vx| = |Vy|.
For two vectors to cancel each other (sum to zero), they must be equal in magnitude and opposite in direction: \mathbf{A} = -\mathbf{B} \quad\Rightarrow\quad \mathbf{A} + \mathbf{B} = \mathbf{0}.

Common Projectile Questions (concept checks)

If you flip a coin in a car moving at constant speed, the coin lands in your hand (in the car frame) due to inertia and no net horizontal acceleration in the car frame.
If the car accelerates, the coin tends to lag behind, and you may miss catching it due to the horizontal pseudo-forces in the non-inertial frame.
Examples of objects in projectile motion:
- A football kicked into the air
- A rock thrown into the air
- Keys dropped off a cliff
Definition of projectile motion:
- The motion of an object that is launched or thrown and moves under the influence of gravity alone (neglecting air resistance), resulting in a characteristic parabolic path.

Quick Reference Formulas (LaTeX)

Vector magnitude: |oldsymbol{v}| = \sqrt{vx^2 + vy^2}
Components of launch velocity: Vx = Va \cos\theta,\quad Vy = Va \sin\theta
Time to top: t{top} = \dfrac{Va \sin\theta}{g}
Total time of flight (level ground): T = \dfrac{2 V_a \sin\theta}{g}
Range: R = \dfrac{V_a^2 \sin(2\theta)}{g}
Maximum range angle: \theta_{max} = 45^{\circ}
Complementary-angle range equality: R(\theta) = R(90^{\circ}-\theta)$$

Note: Some inconsistencies appeared in the source transcript (e.g., B’s direction in Page 1 vs Page 2). The consistent set used here is A = (3,0) and B = (0,-4), yielding the standard results listed above. If your instructor uses a different coordinate convention, adjust the signs accordingly.