| 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 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 ball of mass \(m\) is released from rest at a distance \(h\) above a frictionless plane inclined at an angle of \(45^\circ\) to the horizontal as shown above. The ball bounces horizontally off the plane at point \(P_1\) with the same speed with which it struck the plane and strikes the plane again at point \(P_2\). In terms of \(g\) and \(h\), determine each of the following quantities:
A car accelerates from rest with an acceleration of \( 3.5 \, \text{m/s}^2 \) for \( 10 \, \text{s} \). After this, it continues at a constant speed for an unknown amount of time. The driver notices a ramp \( 50 \, \text{m} \) ahead and takes \( 0.6 \, \text{s} \) to react. After reacting, the driver hits the brakes, which slow the car with an acceleration of \( 7.2 \, \text{m/s}^2 \). Unfortunately, the driver does not stop in time and goes off the \( 3 \, \text{m} \) high ramp that is angled at \( 27^\circ \).
Three identical rocks are launched with identical speeds from the top of a platform of height \( h_0 \).
Which of the following correctly relates the magnitude \( v_y \) of the vertical component of the velocity of each rock immediately before it hits the ground?
A batter hits a fly ball which leaves the bat \( 0.90 \) \( \text{m} \) above the ground at an angle of \( 61^\circ \) with an initial speed of \( 28 \) \( \text{m/s} \) heading toward centerfield. Ignore air resistance.
A golfer hits a shot to a green that is elevated \(2.80 \, \text{m}\) above the point where the ball is struck. The ball leaves the club at a speed of \(18.9 \, \text{m/s}\) at an angle of \(52.0^\circ\) above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
A plane, 220 meters high, is dropping a supply crate to an island below. It is traveling with a horizontal velocity of 150 m/s. At what horizontal distance must the plane drop the supply crate for it to land on the island? Use \( g = 9.81 \, m/s^2\).
A ball is kicked at a speed of \( v_0 \) at an angle \( \theta \) above the horizontal. The ball travels 25 meters horizontally. If the ball is kicked at \( 2v_0 \), what will the horizontal displacement be?
A baseball is hit high and far across a field. Which of the following statements is true? At the highest point:
\(3.1 \text{ m}\)
\(31.9 \text{ m}\)
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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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