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| Derivation/Formula | Reasoning |
|---|---|
| \[\Delta x_A = v_i t + \tfrac{1}{2}(-g)t^2\] | Ball A is dropped, so \(v_i = 0\) and acceleration is \(-g\). |
| \[\Delta x_A = -\tfrac{1}{2} g t^2\] | Simplify because \(v_i = 0\). |
| \[x_A = H + \Delta x_A\] | Vertical position of Ball A relative to ground after time \(t\). |
| \[x_A = H – \tfrac{1}{2} g t^2\] | Substitute \(\Delta x_A\). |
| \[x_B = v_i t + \tfrac{1}{2}(-g)t^2\] | General kinematics for Ball B, launched upward with \(v_i = 30\,\text{m/s}\). |
| \[x_B = 30t – \tfrac{1}{2} g t^2\] | Insert \(v_i = 30\,\text{m/s}\). |
| \[H – \tfrac{1}{2} g t^2 = 30t – \tfrac{1}{2} g t^2\] | Set \(x_A = x_B\) because the balls pass each other. |
| \[H = 30t\] | Identical \(-\tfrac{1}{2} g t^2\) terms cancel. |
| \[H = 30(2.0)\] | Insert \(t = 2.0\,\text{s}\). |
| \[\boxed{H = 60\,\text{m}}\] | Height of the cliff. |
| Derivation/Formula | Reasoning |
|---|---|
| \[v_{xA} = v_i + (-g)t\] | Final velocity of Ball A after time \(t\); initial speed is zero. |
| \[v_{xA} = -g(2.0)\] | Substitute \(t = 2.0\,\text{s}\). |
| \[v_{xA} = -19.6\,\text{m/s}\] | Downward velocity of Ball A. |
| \[v_{xB} = v_i + (-g)t\] | Final velocity of Ball B using the same kinematic relation. |
| \[v_{xB} = 30 – g(2.0)\] | Insert \(v_i = 30\,\text{m/s}\) and \(t = 2.0\,\text{s}\). |
| \[v_{xB} = 10.4\,\text{m/s}\] | Upward velocity of Ball B. |
| \[\boxed{v_{xA} = -19.6\,\text{m/s}},\quad \boxed{v_{xB} = 10.4\,\text{m/s}}\] | Velocities when the balls pass. |
Just ask: "Help me solve this problem."
A ball is launched horizontally from a height. At the same time, another ball is dropped vertically from the same height. Which hits the ground first?
A boat is rowed directly upriver at a speed of \(2.5 \, \text{m/s}\) relative to the water. Viewers on the shore find that it is moving at only \(0.5 \, \text{m/s}\) relative to the shore. What is the speed of the river? Is it moving with or against the boat?
A ranger in a national park is driving at \( 56 \, \text{km/h} \) when a deer jumps onto the road \( 65 \, \text{m} \) ahead of the vehicle. After a reaction time of \( t \, \text{s} \), the ranger applies the brakes to produce an acceleration of \( -3 \, \text{m/s}^2 \). What is the maximum reaction time allowed if the ranger is to avoid hitting the deer?
A block is projected up a ramp with an initial speed \( v_0 \). It travels along the surface of the ramp with constant acceleration \( a \). Take the positive direction of motion to be up the ramp. If the acceleration vector points opposite the initial velocity vector, which of the following MUST be true?
An object can move upward while having a downward acceleration.
\(60\,\text{m}\)
\(v_{xA} = -19.6\,\text{m/s},\; v_{xB} = 10.4\,\text{m/s}\)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| 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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