| 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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An eagle is flying horizontally at \(6 \, \text{m/s}\) with a fish in its claws. It accidentally drops the fish.
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.
3 clay balls, labeled A, B, and C are launched from the same height at the same speed as shown above. A is launched at \( 30^\circ \) above horizontal, B is launched horizontally, and C is launched \( 30^\circ \) below the horizontal. They all hit the wall (before reaching the ground) in times \( t_A \), \( t_B \), and \( t_C \) respectively. Rank these times from least to greatest.
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 javelin thrower, of height \( 1.8 \) \( \text{m} \), throws a javelin with initial velocity of \( 26 \) \( \text{m s}^{-1} \) at \( 38^{\circ} \) to the horizontal. Calculate the time taken for the javelin to reach the ground from its maximum height. Give your answer in seconds and to an appropriate number of significant figures.
A ball is launched at an angle. At the peak of its trajectory, which of the following is true?
A bald eagle in level flight at a height of \(135 \, \text{m}\) drops the fish it caught. If the eagle’s speed is \(25.0 \, \text{m/s}\) how far from the drop point will the fish land?
A soccer ball is kicked horizontally off an \( 85 \) \( \text{m} \) high cliff at a speed of \( 34 \) \( \text{m/s} \). What is the ball’s final speed when it hits the ground below?
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.
\(30.8\,\text{m}\)
\(5.02\,\text{s}\)
\(1.36\times10^{2}\,\text{m}\)
\(28.9\,\text{m/s}\)
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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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