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# Part (a): What is the difference in the time the balls spend in the air?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | y = v_0 t + \frac{1}{2}gt^2 | Use the kinematic equation for vertical motion to find the time of flight of each ball. |
| 2 | 19.6 = 0 + 14.7t + \frac{1}{2}(9.8)t^2 | For the ball thrown downward: [katex] v_0 = -14.7 \, \text{m/s} [/katex], [katex] g = 9.8 \, \text{m/s}^2 [/katex] |
| 3 | 19.6 = 14.7t + 4.9t^2 \Rightarrow 0 = 4.9t^2 + 14.7t – 19.6 | Solve the quadratic equation using the quadratic formula. |
| 4 | t = \frac{-14.7 \pm \sqrt{(14.7)^2 – 4(4.9)(-19.6)}}{2(4.9)} | The positive root is taken since time cannot be negative. |
| 5 | t = 1 \, \text{s} | Solving the quadratic equation gives [katex] t = 1 \, \text{s} [/katex]. |
| 6 | y = v_0 t + \frac{1}{2}gt^2 | Now for the ball thrown upward: initial speed [katex] v_0 = 14.7 \, \text{m/s} |
| 7 | 0 = 14.7t – \frac{1}{2}(9.8)t^2 \Rightarrow t(2.9 – 2.9t) = 0 \Rightarrow t = 3 \, \text{s} | Solving for \( t [/katex] when the ball reaches the top of its trajectory. |
| 8 | \text{Total time in air} = (2 \times 1 + 3)\ s = 5 \, \text{.} | The time the second ball spends in the air is double the ascent time plus the fall time. |
| 9 | Difference in time = 2 \, \text{s} – 1\,\text{s} = 1s | The difference in the time the balls spend in the air. |
# Part (b): What is the velocity of each ball as it strikes the ground?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | v = v_0 + gt | Use the first kinematic equation to find the final velocity. |
| 2 | v = -14.7 + 9.8 \times 1 = -24.5 \, \text{m/s} | For the ball thrown downward: add gravitational acceleration effect. |
| 3 | v = 14.7 – 9.8 \times 3 = -14.7 \, \text{m/s} | For the ball thrown upward: subtract the effect of gravity over fall time. |
# Part (c): How far apart are the balls 0.800 s after they are thrown?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | d_{downward} = v_0 t + \frac{1}{2}gt^2 | Use kinematic equation for the ball thrown downward. |
| 2 | d_{downward} = 14.7 \times 0.800 + \frac{1}{2} \times 9.8 \times 0.800^2 = 15.68 \, \text{m} | Calculate distance traveled by the downward-thrown ball in 0.800s. |
| 3 | d_{upward} = v_0 t – \frac{1}{2}gt^2 | Use kinematic equation for the ball thrown upward. |
| 4 | d_{upward} = 14.7 \times 0.800 – \frac{1}{2} \times 9.8 \times 0.800^2 = 6.864 \, \text{m} | Calculate distance traveled by the upward-thrown ball in 0.800s. |
| 5 | 20 \, \text{m} | Distance between the balls after 0.800 s is [katex]15.68\, \text{m} + 6.864\, \text{m}[/katex] = 22.544 |
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| Kinematics | Forces |
|---|---|
| [katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |
| [katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |
| [katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |
| [katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |
| Circular Motion | Energy |
|---|---|
| [katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |
| [katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |
| [katex]KE_i + PE_i = KE_f + PE_f[/katex] |
| Momentum | Torque and Rotations |
|---|---|
| [katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |
| [katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |
| [katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |
| Simple Harmonic Motion |
|---|
| [katex]F = -k x[/katex] |
| [katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |
| [katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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