An eagle is flying horizontally at 6.0 m/s with a fish in its claws. It accidentally drops the fish.

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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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^{2} for 10 seconds. After this, it continues at a constant speed for an unknown amount of time. The driver notices a ramp 50 m ahead and takes 0.6 seconds to react. After reacting, the driver hits the brakes which slow the car with an acceleration of 7.2 m/s^{2}. Unfortunately, the driver does not stop in time and goes off the 3m high ramp that is angled at 27°.

A textbook is launched at 20m/s at an angle of 36° from the 12 m high roof.

_{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.

*h* when fired straight up. If the same gun is pointed at an angle of 45° from the vertical, what is the new maximum height of the projectile?

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.

*M* is attached to a string of length *L*. It moves in a vertical circle and at the bottom the ball just clears the ground. The tension at the bottom of the path is 3 times the weight of the ball. Give all answers in terms of *M*, *L*, and *g*.

*t _{A}*,

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?

_{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}.

- 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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