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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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Three identical rocks are launched with identical speeds from the top of a platform of height h0.
Which of the following correctly relates the magnitude vy of the vertical component of the velocity of each rock immediately before it hits the ground?
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The highest barrier that a projectile can clear is 16.2 m, when the projectile is launched at an angle of 22.0° above the horizontal. What is the projectile’s launch speed?
One end of a spring is attached to a solid wall while the other end just reaches to the edge of a horizontal, frictionless tabletop, which is a distance [katex] h [/katex] above the floor. A block of mass M is placed against the end of the spring and pushed toward the wall until the spring has been compressed a distance [katex] x [/katex]. The block is released and strikes the floor a horizontal distance [katex] D [/katex] from the edge of the table. Air resistance is negligible.
Derive an expressions for the following quantities only in terms of [katex] M, x, D, h, [/katex] and any constants.
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 |
|---|---|
| [katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |
| [katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |
| [katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |
| [katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |
| Circular Motion | Energy |
|---|---|
| [katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |
| [katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |
| [katex]KE_i + PE_i = KE_f + PE_f[/katex] |
| Momentum | Torque and Rotations |
|---|---|
| [katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |
| [katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |
| [katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |
| Simple Harmonic Motion |
|---|
| [katex]F = -k x[/katex] |
| [katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |
| [katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |
| 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] |
General Metric Conversion Chart
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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