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

1 | v^2 = u^2 + 2as | Kinematic equation for final velocity (v), with initial velocity (u = 0), acceleration (a), and displacement (s). |

2 | d = v_{0x} t | Horizontal range equation with d as distance and t as time. |

3 | v_{0x} = v_0 \cos(\theta) | Horizontal velocity component in terms of release velocity (v_0) and angle (\theta). |

4 | t = \frac{d}{v_0 \cos(\theta)} | Time of flight in terms of horizontal range and horizontal velocity component. |

5 | v_{0y} = v_0 \sin(\theta) | Vertical velocity component in terms of release velocity and angle. |

6 | 0 = v_{0y} t – \frac{1}{2} g t^2 | Vertical motion equation with v_{0y} as initial vertical velocity and g as gravitational acceleration. |

7 | t = \frac{2v_{0y}}{g} | Solve for time of flight t. |

8 | v_0 = \frac{d}{t \cos(\theta)} | Substitute t into horizontal range equation to solve for v_0. |

9 | a = \frac{v_0^2}{2s} | Use the found value of v_0 to solve for acceleration a. |

10 | v_0 = 23.98 , \text{m/s}, t = 2.60 , \text{s}, a = 410.58 , \text{m/s}^2 | Final values calculated from the equations above. |

Acceleration: \boxed{a = 410.58 , \text{m/s}^2}

Time: \boxed{t = 2.60 , \text{s}}

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- Statistics

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

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

Mathematical

FRQ

An airplane with a speed of 97.5 m/s is climbing upward at an angle of 50.0° with respect to the horizontal. When the plane’s altitude is 732 m, the pilot releases a package.

- Projectiles

Intermediate

Mathematical

GQ

A arrow is shot horizontally from a distance of 20 meters away. It lands .05 meters below the center of the target. If air resistance is negligible what was the initial speed of the arrow?

- Projectiles

Advanced

Mathematical

FRQ

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 h 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 x . The block is released and strikes the floor a horizontal distance D from the edge of the table. Air resistance is negligible.

Derive an expressions for the following quantities only in terms of M, x, D, h, and any constants.

- Energy, Projectiles

- 436.9 m/s
^{2} - 2.6 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_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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