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(a) How far does the textbook travel horizontally after it is released?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v_{0x} = v_0 \cos(\theta)[/katex] | Calculate the initial horizontal velocity. Use the initial speed and the angle of projection. |
| 2 | [katex]v_{0y} = v_0 \sin(\theta)[/katex] | Calculate the initial vertical velocity. Use the initial speed and the angle of projection. |
| 3 | [katex]v_{0x} = 20 \cos(36^\circ) \approx 16.2 \, \text{m/s}[/katex] | Substitute [katex] v_0 = 20 \, \text{m/s} [/katex] and [katex] \theta = 36^\circ [/katex] into the horizontal velocity formula. |
| 4 | [katex]v_{0y} = 20 \sin(36^\circ) \approx 11.8 \, \text{m/s}[/katex] | Substitute [katex] v_0 = 20 \, \text{m/s} [/katex] and [katex] \theta = 36^\circ [/katex] into the vertical velocity formula. |
| 5 | [katex]y = v_{0y} t – \frac{1}{2} g t^2 + \text{initial height}[/katex] | Use the equation of motion in the vertical direction. The textbook is moving under gravity. |
| 6 | [katex]0 = 12 + 11.8 t – \frac{1}{2} \cdot 9.8 t^2[/katex] | Set the displacement [katex] y [/katex] to zero because we are calculating the time [katex] t [/katex] when the textbook reaches the ground. [katex] g = 9.8 \text{m/s}^2 [/katex]. |
| 7 | [katex]4.9 t^2 – 11.8 t – 12 = 0[/katex] | Simplify the quadratic equation to solve for [katex] t [/katex]. |
| 8 | [katex] t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} [/katex] | Use the quadratic formula where [katex] a = 4.9 [/katex], [katex] b = -11.8 [/katex], and [katex] c = -12 [/katex]. |
| 9 | [katex] t \approx 3.18 \, \text{s} [/katex] | Solve the equation and take the positive root. This is the time the textbook stays in the air. |
| 10 | [katex] x = v_{0x} \cdot t [/katex] | Calculate the horizontal distance the textbook travels. Use the horizontal velocity and the time. |
| 11 | [katex] x \approx 16.2 \times 3.18 \approx 51.5 \, \text{m} [/katex] | Substitute [katex] v_{0x} = 16.2 \, \text{m/s} [/katex] and [katex] t = 3.18 \, \text{s} [/katex] into the horizontal distance formula to get the final answer. |
| [katex] \text{The horizontal distance traveled is approximately } 51.5 \, \text{m}[/katex] | ||
(b) What is the book’s velocity (speed and direction) when it reaches the ground?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v_y = v_{0y} – g t [/katex] | Calculate the final vertical velocity using the initial vertical velocity, gravitational acceleration, and time. |
| 2 | [katex]v_y = 11.8 – 9.8 \times 3.18 \approx -19.4 \, \text{m/s}[/katex] | Substitute [katex] v_{0y} = 11.8 \, \text{m/s} [/katex], [katex] g = 9.8 \, \text{m/s}^2 [/katex], and [katex] t = 3.18 \, \text{s} [/katex] into the vertical velocity formula. |
| 3 | [katex]v_{\text{total}} = \sqrt{v_{x}^2 + v_y^2} [/katex] | Calculate the magnitude of the total velocity using the Pythagorean theorem. |
| 4 | [katex]v_{\text{total}} \approx \sqrt{16.2^2 + (-19.4)^2} \approx 25.3 \, \text{m/s} [/katex] | Substitute [katex] v_{x} = 16.2 \, \text{m/s} [/katex] and [katex] v_y = -19.4 \, \text{m/s} [/katex] into the total velocity formula. |
| 5 | [katex]\theta = \tan^{-1} \left(\frac{v_y}{v_x}\right) [/katex] | Calculate the direction of the velocity. Use the inverse tangent to find the angle. |
| 6 | [katex]\theta \approx \tan^{-1} \left(\frac{-19.4}{16.2}\right) \approx -50.1^\circ [/katex] | The vector points 50.1° below the x-axis. |
| [katex] \text{The velocity when the book reaches the ground is approximately } 25.3 \, \text{m/s} \text{ at } -50.1^\circ [/katex] | ||
(c) What is the book’s maximum height above the ground?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v_y = 0 [/katex] | The vertical velocity at the maximum height is zero. |
| 2 | [katex]v_y = v_{0y} – g t[/katex] | Use the vertical motion equation to find the time to reach maximum height. |
| 3 | [katex]0 = 11.8 – 9.8 t[/katex] | Set final vertical velocity [katex] v_y = 0 [/katex] and solve for [katex] t [/katex]. |
| 4 | [katex]t = \frac{11.8}{9.8} \approx 1.20 \, \text{s}[/katex] | Solving the equation gives the time to reach maximum height. |
| 5 | [katex]H = v_{0y} t – \frac{1}{2} g t^2 + \text{initial height}[/katex] | Use the vertical motion equation to find the maximum height. |
| 6 | [katex]H \approx 11.8 \times 1.2 – \frac{1}{2} \times 9.8 \times (1.2)^2 + 12 [/katex] | Substitute [katex] v_{0y} = 11.8 \, \text{m/s} [/katex], [katex] g = 9.8 \, \text{m/s}^2 [/katex], [katex] t = 1.2 \, \text{s} [/katex], and initial height = 12 m. |
| 7 | [katex]H \approx 19.1 \, \text{m}[/katex] | Calculate the maximum height above the ground. |
| [katex] \text{The maximum height above the ground is approximately } 19.1 \, \text{m} [/katex] | ||
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A block of mass \(M_1\) travels horizontally with a constant speed \(v_0\) on a plateau of height \(H\) until it comes to a cliff. A toboggan of mass \(M_2\) is positioned on level ground below the cliff. The center of the toboggan is a distance \(D\) from the base of the cliff.
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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.
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a) 51.5 m
b) 25.3 m/s at 50.1° below the horizontal
c) 19.1 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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