| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[v_{iy}=v_i\sin\theta\] | Resolve the initial speed into its vertical component \(v_{iy}\). |
| 2 | \[0 = v_{iy}^2 + 2(-g)\Delta y\] | At the peak the vertical velocity is zero; apply the kinematic equation with acceleration \(-g\). |
| 3 | \[\Delta y = \frac{v_{iy}^2}{2g}\] | Solve algebraically for the vertical displacement \(\Delta y\), the maximum height. |
| 4 | \[v_{iy}=36.6\sin42.2^\circ = 24.58\,\text{m/s}\] | Insert the given numbers to get \(v_{iy}\). |
| 5 | \[h_{\text{max}} = \frac{(24.6\,\text{m/s})^2}{2(9.80\,\text{m/s}^2)} = 30.83\,\text{m}\] | Calculate the numerical value of the height. |
| 6 | \[\boxed{30.8\,\text{m}}\] | Maximum height reached. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[t = \frac{2v_{iy}}{g}\] | Round-trip time is twice the time to reach the peak, using symmetry of the motion. |
| 2 | \[t = \frac{2(24.6\,\text{m/s})}{9.80\,\text{m/s}^2} = 5.02\,\text{s}\] | Substitute \(v_{iy}\) and \(g\). |
| 3 | \[\boxed{5.02\,\text{s}}\] | Total time in the air. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[v_x = v_i\cos\theta\] | Resolve the initial speed into its horizontal component \(v_x\). |
| 2 | \[v_x = 36.6\cos42.2^\circ = 27.1\,\text{m/s}\] | Insert the given numbers. |
| 3 | \[R = v_x t\] | The horizontal distance equals horizontal speed times total time (no horizontal acceleration). |
| 4 | \[R = 27.1\,\text{m/s}\times5.02\,\text{s} = 1.36\times10^{2}\,\text{m}\] | Compute the range. |
| 5 | \[\boxed{1.36\times10^{2}\,\text{m}}\] | Total horizontal distance. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[v_x = 27.1\,\text{m/s}\] | Horizontal speed remains constant throughout the flight. |
| 2 | \[v_y = v_{iy} – g t\] | Use the kinematic relation for vertical velocity after time \(t\). |
| 3 | \[v_y = 24.6\,\text{m/s} – (9.80\,\text{m/s}^2)(1.50\,\text{s}) = 9.9\,\text{m/s}\] | Insert the numbers to find \(v_y\) at \(1.50\ ~\text{s}\). |
| 4 | \[v = \sqrt{v_x^2 + v_y^2}\] | Speed is the magnitude of the velocity vector. |
| 5 | \[v = \sqrt{(27.1\,\text{m/s})^2 + (9.9\,\text{m/s})^2} = 28.9\,\text{m/s}\] | Compute the magnitude. |
| 6 | \[\boxed{28.9\,\text{m/s}}\] | Speed \(1.50\ ~\text{s}\) after launch. |
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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.
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?
A gun can fire a bullet to height \( h \) when fired straight up. If the same gun is pointed at an angle of \( 45^\circ \) from the vertical, what is the new maximum height of the projectile?
Wile E. Coyote is (still) chasing after his arch-nemesis, the Roadrunner across a cliff that is \(125 \, \text{m}\) high. The Coyote is running in the horizontal direction towards the edge of a cliff when, at the last second, the Roadrunner steps out of the way and the witless coyote falls to the canyon floor.
A baseball is thrown at an angle of 25° relative to the ground at a speed of 23.0 m/s. The ball is caught 42.0 m from the thrower.
A circus cannon fires an acrobat into the air at an angle of \( 45^\circ \) above the horizontal, and the acrobat reaches a maximum height \( y \) above her original launch height. The cannon is now aimed so that it fires straight up, at an identical speed, into the air at an angle of \( 90^\circ \) to the horizontal. In terms of \( y \), what is the acrobat’s new maximum height?
Which statements are not valid for a projectile? Take up as positive.
A ball is launched at an angle. At the peak of its trajectory, which of the following is true?
A ball of mass \( M \) is attached to a string of length \( L \). It moves in a vertical circle and at the bottom the ball just clears the ground. The tension at the bottom of the path is \( 3 \) times the weight of the ball. Give all answers in terms of \( M \), \( L \), and \( g \).
If a baseball pitch leaves the pitcher’s hand horizontally at a velocity of \( 150 \) \( \text{km/h} \), by what \( \% \) will the pull of gravity change the magnitude of the velocity when the ball reaches the batter, \( 18 \) \( \text{m} \) away? For this estimate, ignore air resistance and spin on the ball.
\(30.8\,\text{m}\)
\(5.02\,\text{s}\)
\(1.36\times10^{2}\,\text{m}\)
\(28.9\,\text{m/s}\)
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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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