Step | Formula / Calculation | Reasoning |
---|---|---|

1 | \text{Total initial speed} = \sqrt{v_{\text{initial}}^2 + v_{\text{vertical initial}}^2} | Calculating the total initial speed of the fish combining its horizontal and vertical velocities. |

2 | \text{Total initial speed} = \sqrt{6.0^2 + 0^2} = 6.0 , \text{m/s} | Initial vertical velocity is zero as the fish is dropped, not thrown downwards. |

3 | v_{\text{vertical double}} = \sqrt{\text{double speed}^2 – v_{\text{initial}}^2} | Calculating the vertical component of velocity when the total speed doubles. |

4 | \text{Double speed} = 2 \times \text{total initial speed} = 12.0 , \text{m/s} | Determining the speed value that is double the initial speed. |

5 | v_{\text{vertical double}} = \sqrt{12.0^2 – 6.0^2} | Finding the vertical component when total speed is 12.0 m/s. |

6 | t_{\text{double}} = \frac{v_{\text{vertical double}} – v_{\text{vertical initial}}}{g} | Time calculation for the speed to double using the kinematic equation. |

7 | t_{\text{double}} \approx 1.06 s | Evaluating the time for the fish’s speed to double. |

8 | v_{\text{vertical quadruple}} = \sqrt{\text{quadruple speed}^2 – v_{\text{initial}}^2} | Calculating the vertical component of velocity when the total speed quadruples. |

9 | \text{Quadruple speed} = 2 \times \text{double speed} = 24.0 , \text{m/s} | Determining the speed value that is double the double speed. |

10 | v_{\text{vertical quadruple}} = \sqrt{24.0^2 – 6.0^2} | Finding the vertical component when total speed is 24.0 m/s. |

11 | t_{\text{quadruple}} = \frac{v_{\text{vertical quadruple}} – v_{\text{vertical initial}}}{g} | Time calculation for the speed to quadruple. |

12 | t_{\text{quadruple}} \approx 2.37 s | Evaluating the time for the fish’s speed to quadruple. |

13 | \text{Additional time} = t_{\text{quadruple}} – t_{\text{double}} | Calculating the additional time required for the fish’s speed to double again. |

14 | \text{Additional time} \approx 1.31 s | Evaluating the additional time needed for the fish’s speed to double again from the double speed. |

The time for the fish’s speed to double is approximately 1.06 seconds. The additional time required for the fish’s speed to double again (from double to quadruple the initial speed) is approximately 1.31 seconds.

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- Statistics

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

Advanced

Mathematical

GQ

A marble is thrown horizontally with a speed of 15 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of 65° with the horizontal. From what height above the ground was the marble thrown?

- Projectiles

Intermediate

Proportional Analysis

GQ

- Projectiles

Advanced

Mathematical

GQ

A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.0° above the horizontal. The rocket is fired toward an 11.0-m-high wall, which is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?

- 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

- 1.06 s
- 1.31 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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