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| Derivation/Formula | Reasoning |
|---|---|
| \[R = \frac{2v_i \cos\theta\; v_i \sin\theta}{g} = \frac{v_i^2 \sin 2\theta}{g}\] | The horizontal range on level ground depends on the initial speed \(v_i\), launch angle \(\theta\), and gravity \(g\). |
| \[\sin 2\theta = \frac{Rg}{v_i^2}\] | Solve previous expression for \(\sin 2\theta\) using \(R = 35\,\text{m}\) and \(v_i = 20\,\text{m/s}\). |
| \[v_{ix}^2 + v_{iy}^2 = v_i^2\qquad 2 v_{ix} v_{iy} = Rg\] | Relate horizontal (\(v_{ix}\)) and vertical (\(v_{iy}\)) components to the given speed and range. |
| \[v_{ix} = 9.85\,\text{m/s}, \quad v_{iy} = 17.4\,\text{m/s}\] | Solving the system yields two possible pairs; only this pair lets the height exceed \(5\,\text{m}\), so it describes the physical trajectory. |
| \[y = v_{iy} t – \tfrac{1}{2} g t^2\] | Vertical position after time \(t\) for constant downward acceleration. |
| \[4.9 t^2 – v_{iy} t + 5 = 0\] | Set \(y = 5\,\text{m}\) and substitute \(v_{iy} = 17.4\,\text{m/s}\). |
| \[t = \frac{v_{iy} \pm \sqrt{v_{iy}^2 – 4(4.9)(5)}}{9.8} = 0.315\,\text{s},\; 3.24\,\text{s}\] | Quadratic formula gives the two instants when the projectile is \(5\,\text{m}\) high (ascending and descending). |
| \[x = v_{ix} t\] | Horizontal motion has constant speed \(v_{ix}\), so position equals speed times time. |
| \[\boxed{x \approx 3.1\,\text{m}\quad\text{and}\quad 32\,\text{m}}\] | Multiply \(v_{ix} = 9.85\,\text{m/s}\) by each time to obtain the two horizontal positions. |
Just ask: "Help me solve this problem."
In the absence of air resistance, a projectile is launched from and returns to ground level and has a range of \( 23 \, \text{m} \). Suppose the launch speed is doubled, and the projectile is fired at the same angle above the ground. What is the new range?
A baseball rolls off a 0.70 m high desk and strikes the floor 0.25 m away from the base of the desk. How fast was the ball rolling?
A baseball is hit high and far. Which of the following statements is true? At the highest point:
A soccer ball with an initial height of \(1.5 \, \text{m}\) above the ground is launched at an angle of \(30^\circ\) above the horizontal. The soccer ball travels a horizontal distance of \(45 \, \text{m}\) to a \(9.0 \, \text{m}\) high castle wall, and passes over \(3.20 \, \text{m}\) above the highest point of the wall. Assume air resistance is negligible.
Two balls are launched at the same time from opposite sides of a \( 100 \) \( \text{m} \) wide canyon. Ball A is launched at \( 20 \) \( \text{m/s} \) at \( 45^{\circ} \) from the left side. Ball B is launched at \( 20 \) \( \text{m/s} \) at \( 45^{\circ} \) from the right side.
\(3.1\text{ m}\)
\(32\text{ m}\)
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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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