| Derivation / Formula | Reasoning |
|---|---|
| \[x(t) = v_i \cos\theta\, t\] | Horizontal motion is uniform (no horizontal acceleration), so position grows linearly with constant speed \(v_i\cos\theta\). |
| \[y(t) = v_i \sin\theta \, t – \tfrac{1}{2} g t^2\] | Vertical motion has initial upward speed \(v_i\sin\theta\) and constant downward acceleration \(g\), giving a parabola. |
| \[R=\frac{v_i^2\sin 2\theta}{g}\] | Level-ground range formula (derive by eliminating \(t_f\) from \(x=v_i\cos\theta\,t_f\) and \(0=v_i\sin\theta\,t_f-\tfrac{1}{2}gt_f^2\)). |
| \[\sin 2\theta=\frac{Rg}{v_i^2}=\frac{35\cdot 9.8}{20^2}=0.8575\] | Numerically \(2\theta\approx 59.0^\circ\) or \(121.0^\circ\), so \(\theta\approx 29.5^\circ\) (low) or \(\theta\approx 60.5^\circ\) (high). |
| \[h_{\max}=\frac{(v_i\sin\theta)^2}{2g}\] | Peak height test selects the physically relevant branch for \(y=5\,\text{m}\). |
| \[\begin{aligned} \theta&\approx 29.5^\circ:& h_{\max}&=\frac{(20\sin29.5^\circ)^2}{2\cdot 9.8}\approx 4.95\,\text{m}<5\\ \theta&\approx 60.5^\circ:& h_{\max}&=\frac{(20\sin60.5^\circ)^2}{2\cdot 9.8}\approx 15.46\,\text{m}>5 \end{aligned}\] | The low arc never reaches \(5\,\text{m}\); only the high arc can cross \(y=5\,\text{m}\) twice (upward and downward). |
| \[t=\frac{x}{v_i\cos\theta}\] | From \(x(t)=v_i\cos\theta\,t\), solve for time at a given horizontal position. |
| \[y(x)=v_i\sin\theta\Big(\frac{x}{v_i\cos\theta}\Big)-\frac{1}{2}g\Big(\frac{x}{v_i\cos\theta}\Big)^2\] | Substitute \(t\) into \(y(t)\) to express height directly in terms of \(x\). |
| \[y(x)=x\tan\theta-\frac{g\,x^2}{2v_i^{2}\cos^{2}\theta}\] | Algebraic simplification: \(\tan\theta=\frac{\sin\theta}{\cos\theta}\). |
| \[5=x\tan\theta-\frac{g\,x^2}{2v_i^{2}\cos^{2}\theta}\] | Impose the target height \(y=5\,\text{m}\) on the high-angle trajectory. |
| \[-\underbrace{\frac{g}{2v_i^{2}\cos^{2}\theta}}_{\displaystyle A}\,x^2+\underbrace{\tan\theta}_{\displaystyle B}\,x-\underbrace{5}_{\displaystyle C}=0\] | Identify quadratic coefficients \(a=-A,\; b=B,\; c=-5\). This makes the upcoming plug-in transparent. |
| \[\cos\theta\approx 0.4924,\quad \tan\theta\approx 1.7675,\quad A=\frac{9.8}{2\cdot 20^2\cos^2\theta}\approx 0.05052\] | Using \(\theta\approx 60.5^\circ\). Thus the quadratic is \(-0.05052\,x^2+1.7675\,x-5=0\). |
| \[\Delta=b^2-4ac=1.7675^2-4(-0.05052)(-5)\approx 2.114>0\] | Positive discriminant ⇒ two distinct horizontal positions reach \(y=5\,\text{m}\). |
| \[x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-1.7675\pm \sqrt{2.114}}{2(-0.05052)}\] | Quadratic formula with \(a=-0.05052,\; b=1.7675,\; c=-5\). |
| \[x\approx 3.10\,\text{m}\quad\text{or}\quad x\approx 31.88\,\text{m}\] | Two crossings of the \(5\,\text{m}\) level: once ascending, once descending. |
| \[\boxed{x=3.1\,\text{m},\;31.9\,\text{m}}\] | Final answer, rounded. |
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A ball is tossed straight up while the thrower is standing in a moving train car that is moving at a constant velocity. Neglecting air resistance, what is the path of the ball relative to the ground outside the train?
Two balls are launched at the same time from opposite sides of a \( 100 \) \( \text{m} \) wide and \(1000 ~\text{m}\) 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.
A ball is launched and lands \( 20 \) \( \text{m} \) away below the launch point \( 2.5 \) \( \text{s} \) later. The maximum height reached is \( 8.0 \) \( \text{m} \). What was the original launch velocity?
A bird, traveling at \(50 \, \text{m/s}\) wants to hit a man \(100 \, \text{m}\) below with a dropping. How far in distance before flying directly over the man should the bird release it?
A rocket-powered hockey puck has a thrust of \(4.40 \, \text{N}\) and a total mass of \(1.00 \, \text{kg}\). It is released from rest on a frictionless table, \(2.10 \, \text{m}\) from the edge of a \(2.10 \, \text{m}\) drop. The front of the rocket is pointed directly toward the edge. Assuming that the thrust of the rocket is present for the entire time of travel, how far does the puck land from the base of the table?
The highest barrier that a projectile can clear is 16.2 m, when the projectile is launched at an angle of 22.0° above the horizontal. What is the projectile’s launch speed?
An airplane with a speed of \( 97.5 \, \text{m/s} \) is climbing upward at an angle of \( 50.0^\circ \) with respect to the horizontal. When the plane’s altitude is \( 732 \, \text{m} \), the pilot releases a package.
A cannon fires projectiles on a flat range at a fixed speed but with variable angle. The maximum range of the cannon is \(L\). What is the range of the cannon when it fires at an angle of \(30^\circ\) above the horizontal? Ignore air resistance.
A projectile is launched at an angle of \( 30^{\circ} \) and hits a vertical wall \( 40 \) \( \text{m} \) away. After bouncing back horizontally, it lands \( 15 \) \( \text{m} \) behind the launch point. How high up on the wall did the projectile strike?
A block of mass \(M_1\) travels horizontally with a constant speed \(v_0\) on a plateau of height \(H\) until it comes to a cliff. A toboggan of mass \(M_2\) is positioned on level ground below the cliff. The center of the toboggan is a distance \(D\) from the base of the cliff.
\(3.1 \text{ m}\)
\(31.9 \text{ m}\)
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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] |
Metric Prefixes
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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