(a) How far does the textbook travel horizontally after it is released?

Step | Derivation/Formula | Reasoning |
---|---|---|

1 | v_{0x} = v_0 \cos(\theta) | Calculate the initial horizontal velocity. Use the initial speed and the angle of projection. |

2 | v_{0y} = v_0 \sin(\theta) | Calculate the initial vertical velocity. Use the initial speed and the angle of projection. |

3 | v_{0x} = 20 \cos(36^\circ) \approx 16.2 \, \text{m/s} | Substitute v_0 = 20 \, \text{m/s} and \theta = 36^\circ into the horizontal velocity formula. |

4 | v_{0y} = 20 \sin(36^\circ) \approx 11.8 \, \text{m/s} | Substitute v_0 = 20 \, \text{m/s} and \theta = 36^\circ into the vertical velocity formula. |

5 | y = v_{0y} t – \frac{1}{2} g t^2 + \text{initial height} | Use the equation of motion in the vertical direction. The textbook is moving under gravity. |

6 | 0 = 12 + 11.8 t – \frac{1}{2} \cdot 9.8 t^2 | Set the displacement y to zero because we are calculating the time t when the textbook reaches the ground. g = 9.8 \text{m/s}^2 . |

7 | 4.9 t^2 – 11.8 t – 12 = 0 | Simplify the quadratic equation to solve for t . |

8 | t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} | Use the quadratic formula where a = 4.9 , b = -11.8 , and c = -12 . |

9 | t \approx 3.18 \, \text{s} | Solve the equation and take the positive root. This is the time the textbook stays in the air. |

10 | x = v_{0x} \cdot t | Calculate the horizontal distance the textbook travels. Use the horizontal velocity and the time. |

11 | x \approx 16.2 \times 3.18 \approx 51.5 \, \text{m} | Substitute v_{0x} = 16.2 \, \text{m/s} and t = 3.18 \, \text{s} into the horizontal distance formula to get the final answer. |

\text{The horizontal distance traveled is approximately } 51.5 \, \text{m} |

(b) What is the book’s velocity (speed and direction) when it reaches the ground?

Step | Derivation/Formula | Reasoning |
---|---|---|

1 | v_y = v_{0y} – g t | Calculate the final vertical velocity using the initial vertical velocity, gravitational acceleration, and time. |

2 | v_y = 11.8 – 9.8 \times 3.18 \approx -19.4 \, \text{m/s} | Substitute v_{0y} = 11.8 \, \text{m/s} , g = 9.8 \, \text{m/s}^2 , and t = 3.18 \, \text{s} into the vertical velocity formula. |

3 | v_{\text{total}} = \sqrt{v_{x}^2 + v_y^2} | Calculate the magnitude of the total velocity using the Pythagorean theorem. |

4 | v_{\text{total}} \approx \sqrt{16.2^2 + (-19.4)^2} \approx 25.3 \, \text{m/s} | Substitute v_{x} = 16.2 \, \text{m/s} and v_y = -19.4 \, \text{m/s} into the total velocity formula. |

5 | \theta = \tan^{-1} \left(\frac{v_y}{v_x}\right) | Calculate the direction of the velocity. Use the inverse tangent to find the angle. |

6 | \theta \approx \tan^{-1} \left(\frac{-19.4}{16.2}\right) \approx -50.1^\circ | The vector points 50.1° below the x-axis. |

\text{The velocity when the book reaches the ground is approximately } 25.3 \, \text{m/s} \text{ at } -50.1^\circ |

(c) What is the book’s maximum height above the ground?

Step | Derivation/Formula | Reasoning |
---|---|---|

1 | v_y = 0 | The vertical velocity at the maximum height is zero. |

2 | v_y = v_{0y} – g t | Use the vertical motion equation to find the time to reach maximum height. |

3 | 0 = 11.8 – 9.8 t | Set final vertical velocity v_y = 0 and solve for t . |

4 | t = \frac{11.8}{9.8} \approx 1.20 \, \text{s} | Solving the equation gives the time to reach maximum height. |

5 | H = v_{0y} t – \frac{1}{2} g t^2 + \text{initial height} | Use the vertical motion equation to find the maximum height. |

6 | H \approx 11.8 \times 1.2 – \frac{1}{2} \times 9.8 \times (1.2)^2 + 12 | Substitute v_{0y} = 11.8 \, \text{m/s} , g = 9.8 \, \text{m/s}^2 , t = 1.2 \, \text{s} , and initial height = 12 m. |

7 | H \approx 19.1 \, \text{m} | Calculate the maximum height above the ground. |

\text{The maximum height above the ground is approximately } 19.1 \, \text{m} |

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- Statistics

Advanced

Conceptual

MCQ

Three identical rocks are launched with identical speeds from the top of a platform of height h_{0}.

- Rock 1 is launched at a 45° angle above the horizontal
- Rock 2 is launched at a 45° angle below the horizontal
- Rock 3 is launched horizontally

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?

- Projectiles

Advanced

Proportional Analysis

GQ

A rifle is used to shoot a target twice, using identical cartridges. The first time, the rifle is aimed parallel to the ground and directly at the center of the bull’s-eye. The bullet strikes the target at a distance of H_{A} below the center, however. The second time, the rifle is similarly aimed, but from twice the distance from the target. This time the bullet strikes the target at a distance of H_{B} below the center. Find the ratio Н_{B}/ Н_{А}.

- Projectiles

Advanced

Mathematical

GQ

- Projectiles

Intermediate

Conceptual

MCQ

A golfer hits her ball in a high arcing shot. Air resistance is negligible. When the ball is at its highest point, which of the following is true?

- Projectiles

Intermediate

Mathematical

GQ

A major-league pitcher can throw a baseball in excess of 41.0 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0 m away from the point of release?

- Projectiles

a) 51.5 m

b) 25.3 m/s at 50.1° below the horizontal

c) 19.1 m

Kinematics | Forces |
---|---|

\Delta x = v_i t + \frac{1}{2} at^2 | F = ma |

v = v_i + at | F_g = \frac{G m_1m_2}{r^2} |

a = \frac{\Delta v}{\Delta t} | f = \mu N |

R = \frac{v_i^2 \sin(2\theta)}{g} |

Circular Motion | Energy |
---|---|

F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |

a_c = \frac{v^2}{r} | PE = mgh |

KE_i + PE_i = KE_f + PE_f |

Momentum | Torque and Rotations |
---|---|

p = m v | \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 |
---|

F = -k x |

T = 2\pi \sqrt{\frac{l}{g}} |

T = 2\pi \sqrt{\frac{m}{k}} |

Constant | Description |
---|---|

g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |

G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |

\mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |

k | Spring constant, in \text{N/m} |

M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |

M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |

M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |

Variable | SI Unit |
---|---|

s (Displacement) | \text{meters (m)} |

v (Velocity) | \text{meters per second (m/s)} |

a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |

t (Time) | \text{seconds (s)} |

m (Mass) | \text{kilograms (kg)} |

Variable | Derived SI Unit |
---|---|

F (Force) | \text{newtons (N)} |

E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |

P (Power) | \text{watts (W)} |

p (Momentum) | \text{kilogram meters per second (kgm/s)} |

\omega (Angular Velocity) | \text{radians per second (rad/s)} |

\tau (Torque) | \text{newton meters (Nm)} |

I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |

f (Frequency) | \text{hertz (Hz)} |

General Metric Conversion Chart

Conversion Example

Example of using unit analysis: Convert 5 kilometers to millimeters.

Start with the given measurement:

`\text{5 km}`

Use the conversion factors for kilometers to meters and meters to millimeters:

`\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}`

Perform the multiplication:

`\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}`

Simplify to get the final answer:

`\boxed{5 \times 10^6 \, \text{mm}}`

Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|

Pico- | p | 10^{-12} | 0.000000000001 |

Nano- | n | 10^{-9} | 0.000000001 |

Micro- | µ | 10^{-6} | 0.000001 |

Milli- | m | 10^{-3} | 0.001 |

Centi- | c | 10^{-2} | 0.01 |

Deci- | d | 10^{-1} | 0.1 |

(Base unit) | – | 10^{0} | 1 |

Deca- or Deka- | da | 10^{1} | 10 |

Hecto- | h | 10^{2} | 100 |

Kilo- | k | 10^{3} | 1,000 |

Mega- | M | 10^{6} | 1,000,000 |

Giga- | G | 10^{9} | 1,000,000,000 |

Tera- | T | 10^{12} | 1,000,000,000,000 |

- Some answers may be slightly off by 1% depending on rounding, etc.
- Answers will use different values of gravity. Some answers use 9.81 m/s
^{2}, and other 10 m/s^{2 }for calculations. - Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
- Bookmark questions that you can’t solve so you can come back to them later.
- Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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