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| 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?
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:
Two balls are launched at the same speed. Ball A is launched at an angle of \( 45^{\circ} \) and Ball B is launched at an angle of \( 60^{\circ} \). Which one reaches a higher point?
An arrow is shot horizontally from a distance of \( 20 \, \text{m} \) away. It lands \( 0.05 \, \text{m} \) below the center of the target. If air resistance is negligible, what was the initial speed of the arrow?
A ball is thrown horizontally from the roof of a building \( 7.5 \) \( \text{m} \) tall and lands \( 9.5 \) \( \text{m} \) from the base. What was the ball’s initial speed?
Which launch angle gives the greatest horizontal range, assuming level ground and no air resistance?
A ball is shot from the top of a building with an initial velocity of \( 18 \) \( \text{m/s} \) at an angle \( \theta = 42^\circ \) above the horizontal.
A marble is thrown horizontally with a speed of \(15 \, \text{m/s}\) from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of \(65^\circ\) with the horizontal. From what height above the ground was the marble thrown?
A ball of mass \( 0.5 \, \text{kg} \), initially at rest, is kicked directly toward a fence from a point \( 32 \, \text{m} \) away, as shown above. The velocity of the ball as it leaves the kicker’s foot is \( 20 \, \text{m/s} \) at an angle of \( 37^\circ \) above the horizontal. The top of the fence is \( 2.5 \, \text{m} \) high. The ball hits nothing while in flight and air resistance is negligible.
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?
\(3.1 \text{ m}\)
\(31.9 \text{ m}\)
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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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