1D & 2D Kinematics - Lecture 5 Notes

1D Kinematics

• Context from Lecture 5: 1D Kinematics (constant acceleration) and introduction to 2D Kinematics (projectile motion).
• Example scenario introduced: hammer dropped near Moon’s surface with ay = −1.63 m/s² (Moon gravity); goal is to find time to hit the ground given initial conditions.

Key concepts in 1D Kinematics

• Constant acceleration problems use the basic kinematic equations:
  • $y = y<em>0 + v</em>{0y} t + \frac{1}{2} a_y t^2$
  • $v<em>y = v</em>{0y} + a_y t$
  • $v<em>y^2 = v</em>{0y}^2 + 2 a_y \Delta y$
  • (Alternate form, when helpful) $\Delta y = v<em>{n} \Delta t \quad \text{where } v</em>n \text{ is an appropriate velocity term}$
• If both x- and y- directions are present, the horizontal motion can be treated separately when ax = 0 (no horizontal acceleration) and ay is constant.
• Zero or negative slope interpretations on graphs:
  • Instantaneous acceleration vs time: ay is constant (horizontal line) if acceleration is constant.
  • Instantaneous velocity vs time: velocity changes linearly with time when ay is constant.
  • Position vs time: position is a quadratic function of time when ay is constant.
• Example interpretation: hammer under constant negative ay causes velocity to decrease linearly with time; the speed can still increase if the object is moving downward or has reversed direction.

Worked setup (1D, from the slides)

• Known quantities in a typical problem:
  • Voy = 0 m/s
  • ay = −1.63 m/s²
  • Yo = initial height (unknown in general; in a specific numerical example, Yo could be 5 ft or other values after conversion)
• Unknown quantity: time t to hit the ground (or another final position).
• Important initial-step check: ensure consistent units (see feet-to-meters conversion example below).

Unit conversion example (feet to meters)

• Common pitfall: mixing feet with meters in a quadratic time problem.
• Conversion used: 1 ft = 0.3048 m (so 5 ft = 5 × 0.3048 m).
• Example progression:
  • If Yo = 5 ft, convert: $y_0 = 5\text{ ft} = 5(0.3048\text{ m}) = 1.524\text{ m}$
  • With initial vy0 = 0 and ay = −1.63 m/s², solve: $y = y<em>0 + v</em>{0y} t + \frac{1}{2} a<em>y t^2 \ 0 = y</em>0 + 0 \cdot t + \frac{1}{2}(-1.63) t^2$
  • This yields a time t = \sqrt{ \frac{2 y0}{|ay|} } = \sqrt{ \frac{2 \times 1.524}{1.63} } \approx 1.37\text{ s} \n
• Solve for t using the generic case with nonzero v0y if provided:
  • For given Δy and v0y, solve the quadratic in t from $\Delta y = v<em>{0y} t + \frac{1}{2} a</em>y t^2$ and pick the physically meaningful (positive) root.
• Solving quadratics:
  • General form: $A t^2 + B t + C = 0$ where $A = \tfrac{1}{2} a<em>y, \; B = v</em>{0y}, \; C = y<em>0 - y</em>{final}$
  • Discriminant: $\Delta = B^2 - 4 A C$
  • Time solutions: $t = \frac{-B \pm \sqrt{\Delta}}{2A}$
  • Always perform a plausibility check; discard negative time if not physically meaningful.

Quick 1D problem recap (from slides)

• Given Yo, Voy, ay, find t.
• Example result when Yo = 5 ft and ay = −1.63 m/s² (after conversion): t ≈ 1.37–1.5 s depending on initial velocity and exact height.
• Takeaway: for constant ay, position is quadratic in t; velocity is linear in t; acceleration is constant.

2D Kinematics (Projectile Motion)

Key concepts in 2D Kinematics

• In the absence of air resistance, motion separates into independent x and y components.
  • The acceleration has only a y-component: ay = −g (≈ −9.8 m/s²). ax = 0.
  • The velocity vector is v = vx i + vy j, with vx constant and vy changing due to ay.
• Initial speed and angle: for a launch speed v0 at angle θ above the horizontal:
  • $v<em>{0x} = v</em>0 \cos\theta$
  • $v<em>{0y} = v</em>0 \sin\theta$
  • Example: v0 = 15 m/s, θ = 60° → $v<em>{0x} = 7.5\ \text{m/s}, \; v</em>{0y} = 15 \sin 60° \approx 12.99 \text{ m/s}$
• Projectile equations (component form):
  • Horizontal motion (ax = 0): $x = x<em>0 + v</em>{0x} t$ and $v<em>x = v</em>{0x}$
  • Vertical motion (ay = −g): $y = y<em>0 + v</em>{0y} t + \frac{1}{2}(-g) t^2$ and $v<em>y = v</em>{0y} - g t$
• Magnitude of velocity: $v^2 = v<em>x^2 + v</em>y^2$. At any time, velocity components give the instantaneous speed and direction.
• Key qualitative points:
  • The cannonball’s velocity vector points in the direction of motion; its x- and y- components align with the axes.
  • Without air resistance, only the y-component of velocity changes with time; the x-component remains constant.
  • The apex occurs when vy = 0; at the apex, the acceleration is still downward (g).
• Mass, size, or shape do not affect the motion in a gravitational field when air resistance is neglected; all objects have the same acceleration in the y-direction (g).

Worked 2D problem (projectile from height with initial speed at angle)

• Given: initial speed v0 = 15 m/s at θ = 60°; thus
  • $v<em>{0x} = 7.5 \text{ m/s}, \quad v</em>{0y} = 15 \sin 60° \approx 12.99 \text{ m/s}$
• The vertical displacement Δy is known (e.g., −3 m if final height is 0 and initial height is 3 m):
  • Solve $\Delta y = v<em>{0y} t + \frac{1}{2} a</em>y t^2$ with $a_y = -9.8 \text{ m/s}^2$
  • For Δy = −3 m and v0y ≈ 13 m/s, the quadratic is: $4.9 t^2 - 13 t - 3 = 0$
  • Solutions: $t = \frac{ -B \pm \sqrt{B^2 - 4AC} }{2A} = \frac{13 \pm \sqrt{((-13)^2) - 4(4.9)(-3)}}{2(4.9)}$
  • Numerically: $t \approx 2.87\text{ s} \quad \text{and} \quad t \approx -0.214\text{ s}$
  • Physically, take the positive root: t ≈ 2.87 s.
• Horizontal displacement (range at that time):
  • Since ax = 0 and x = x0 + v0x t, with x0 = 0: $\Delta x = v_{0x} t = 7.5 \times 2.87 \approx 21.5 \text{ m}$
• Final vertical velocity vy at impact can be found via energy-like relation or kinematics:
  • Energy/kinematics approach: $v<em>y^2 = v</em>{0y}^2 + 2 a_y \Delta y$
  • With v0y ≈ 13 m/s and Δy = −3 m: $v_y^2 = 13^2 + 2(-9.8)(-3) = 169 + 58.8 = 227.8$
  • Thus, $v_y = \pm \sqrt{227.8} \approx \pm 15.1 \text{ m/s}$
  • For the landing (downward direction), take v_y ≈ −15 m/s.
• Time from vertical velocity equation to landing (consistency check):
  • $v<em>y = v</em>{0y} + a_y t \Rightarrow -15 = 12.99 - 9.8 t$
  • Solve: $t \approx \frac{12.99 + 15}{9.8} \approx 2.86 \text{ s}$
• Summary of the steps and logic:
  • Decompose into x and y components using v0 and θ.
  • Use y-equation to solve for flight time t (quadratic in t).
  • Use x-equation (with ax = 0) to compute horizontal range using the t found.
  • Use vy relation to find vertical speed at impact or to confirm time consistency.

Solving a generic 2D time problem (quadratic form)

• In problems where Δy = v0y t + (1/2) a_y t^2, you may get a quadratic in t:
  • $A t^2 + B t + C = 0, \quad A = \tfrac{1}{2} a<em>y, \; B = v</em>{0y}, \; C = y<em>0 - y</em>{final}$
  • Example coefficients from the problem: $A = 4.9, \; B = -13, \; C = -3$
• Quadratic formula:
  • $t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$
• After computing the two roots, always pick the physically meaningful one (usually the positive root). If both are positive, use context to determine which is the correct time (the later crossing might be after impact, etc.).
• A useful check: if the problem uses v_y or y-terms with a downward acceleration, the final angular or distance values should be physically plausible given the launch height and speed.

Important takeaways for 2D/projectile problems

• Independence of axes: horizontal motion does not affect vertical motion when air resistance is neglected.
• Horizontal range is driven by v0x and time of flight: $\Delta x = v_{0x} t$ when ax = 0.
• Time of flight depends on vertical motion; for launch from height y0 to ground level y = 0, solve the vertical equation to obtain t, then plug into the horizontal equation for Δx.
• Apex considerations: vy = 0 at the top; the acceleration remains downward (g) throughout; velocity is purely horizontal at the apex if launched upward.
• In standard projectile problems, the only acceleration is gravity downward, so the horizontal motion remains uniform while vertical motion is uniformly accelerated.

Worked recap of a canonical 2D problem (from slides)

• Given: launch with v0 = 15 m/s at θ = 60°, y0 = 3 m, y(final) = 0, g = 9.8 m/s².
• Computations:
  • Components: $v<em>{0x} = 7.5 \text{ m/s}, \quad v</em>{0y} = 15 \sin 60° \approx 12.99 \text{ m/s}$
  • Vertical motion: solve $\Delta y = v_{0y} t + \frac{1}{2}(-g) t^2 \Rightarrow -3 = 12.99 t - 4.9 t^2$
  • Quadratic form: $4.9 t^2 - 12.99 t - 3 = 0$
  • Positive root yields t ≈ 2.87 s; horizontal range: $\Delta x = v_{0x} t ≈ 7.5 \times 2.87 ≈ 21.5 \text{ m}$
  • Final vy: $v<em>y^2 = v</em>{0y}^2 + 2(-g)\Delta y = 12.99^2 + 2(-9.8)(-3) ≈ 227.8$
  • ⇒ $v<em>y \,≈ \, -15 \text{ m/s}$ at impact; cross-check with $v</em>y = v<em>{0y} + a</em>y t$ gives consistent time.

Additional context from the lecture slides

• Cannonball drop from 10 m takes about 1.43 s to land (t = \sqrt{2h/g} for h = 10 m).
• Horizontally fired projectile lands at the same time as a dropped object from the same height (absence of air resistance); horizontal velocity affects range but not time of fall.
• Mass, size, or shape do not affect the motion in a gravitational field if air resistance is neglected; all objects have the same acceleration due to gravity in the vertical direction.

Quick problem-solving reminders

• Always set up components: choose a coordinate system with up as positive y and right as positive x.
• For horizontal motion with ax = 0, use $x = x<em>0 + v</em>{0x} t$ and $v<em>x = v</em>{0x}$.
• For vertical motion, use $y = y<em>0 + v</em>{0y} t - \frac{1}{2} g t^2$ with g ≈ 9.8 m/s²; or equivalently, ay = −g in the general formulas.
• To find time from vertical motion, solve the quadratic equation in t and choose the positive root.
• To find horizontal range, multiply the time of flight by the horizontal velocity component: $\Delta x = v_{0x} t$.
• Use the relation for velocity magnitude if needed: $v^2 = v<em>x^2 + v</em>y^2$; for vertical velocity at impact, use either vy = v0y − g t or vy^2 = v0y^2 + 2 a_y Δy.
• When in doubt, check units and perform a plausibility check on the sign and magnitude of the results; ensure the time you pick makes physical sense (positive, within the context).

Quick reference equations (summary)

• 1D kinematics (constant ay):

  • $y = y<em>0 + v</em>{0y} t + \frac{1}{2} a_y t^2$
  • $v<em>y = v</em>{0y} + a_y t$
  • $v<em>y^2 = v</em>{0y}^2 + 2 a_y \Delta y$

• 2D kinematics (projectile):

  • $v<em>x = v</em>{0x}, \ x = x<em>0 + v</em>{0x} t$
  • $v<em>y = v</em>{0y} - g t, \ y = y<em>0 + v</em>{0y} t - \frac{1}{2} g t^2$
  • $v^2 = v<em>x^2 + v</em>y^2$
  • $v<em>{0x} = v</em>0 \cos\theta, \quad v<em>{0y} = v</em>0 \sin\theta$

• Quadratic solving:

  • $A t^2 + B t + C = 0, \quad t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

• Special constant-height projectile time (example):

  • Drop from height h: $t = \sqrt{ \frac{2h}{g} }$

• Common constant: 1 ft = 0.3048 m.

• Apex observation: at the apex, vy = 0; acceleration remains downward (g).

Summary takeaway from the lecture

• 1D kinematics teaches how constant acceleration shapes y(t) and v(t); time to fall from height depends on height and acceleration.
• 2D kinematics extends to projectile motion, showing the independence of axes, the role of initial speed and angle, and how to compute time, range, and final velocities using component forms and quadratic solving.
• Realistic checks include unit consistency, choosing physical roots, and recognizing that without air resistance, horizontal motion does not affect vertical timing.