| 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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Projectiles 1 and 2 are launched from level ground at the same time and follow the trajectories shown in the figure. Which one of the projectiles, if either, returns to the ground first, and why?
An eagle is flying horizontally at \(6 \, \text{m/s}\) with a fish in its claws. It accidentally drops the fish.
A stone is thrown horizontally at \(8.0 \, \text{m/s}\) from a cliff \(80 \,\text{m}\) high. How far from the base of the cliff will the stone strike the ground?
A projectile is launched at a speed of \( 22 \) \( \text{m/s} \) at an angle of \( 60^{\circ} \) above the horizontal. It lands on a ramp that is \( 5 \) \( \text{m} \) lower than the launch height. How long does it take for the projectile to hit the ramp?
Consider a ball thrown up from the surface of the earth into the air at an angle of \( 30^\circ \) above the horizontal. Air resistance is negligible. The ball’s acceleration just after release is most nearly
A train is moving to the right at \( 20 \) \( \text{m/s} \). A passenger on the train throws a ball horizontally to the left at \( 5 \) \( \text{m/s} \) (relative to the train).
You kick a soccer ball with an initial velocity directed 53° above the horizontal. The ball lands on a roof 7.2 m high. The wall of the building is 25 m away, and it takes the ball 2.1 seconds to pass directly over the wall.
Suppose the water at the top of Niagara Falls has a horizontal speed of \( 2.7 \, \text{m/s} \) just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a \( 75^\circ \) angle below the horizontal?
A cylindrical tank of water (height \( H \)) is punctured at a height \( h \) above the bottom. How far from the base of the tank will the water stream land (in terms of \( h \) and \( H \))? What must the value of \( h \) be such that the distance at which the stream lands will be equal to \( H \)?
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 [katex] g = 9.81 \, m/s^2[/katex].
\(3.1 \text{ m}\)
\(31.9 \text{ m}\)
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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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