To solve this problem, we can use the principles of kinematics. Specifically, we know the initial velocity in the horizontal direction, the height from which the ball is dropped, and we can calculate the time it takes to hit the ground using the initial vertical velocity and the height. Once we have the time, we can compute the final velocity components in both the horizontal and vertical directions and then calculate the resultant final speed using the Pythagorean theorem.

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

1 | t = \sqrt{\frac{2h}{g}} | Calculate the time to hit the ground. Since the initial vertical velocity (u_y) is 0 (ball is kicked horizontally), the time is solely determined by the height of the cliff (h) and gravity (g), using the equation for free fall: t = \sqrt{\frac{2h}{g}}. |

2 | v_{x} = u_{x} | The horizontal velocity remains constant (denoted as v_x), equal to the initial horizontal velocity (u_x), because there is no acceleration in the horizontal direction. |

3 | v_{y} = gt | Calculate the final vertical velocity (v_y). It is determined only by the acceleration due to gravity and the time in the air, as v_y = g \times t. |

4 | v = \sqrt{v_{x}^2 + v_{y}^2} | Determine the resultant velocity (v) using the Pythagorean theorem, which combines the horizontal (constant) and vertical (accelerated) components of velocity. |

Let’s execute these calculations:

– Given: h = 85 m, u_x = 34 m/s, and g = 9.81 m/s².

1. Time to reach the ground:

t = \sqrt{\frac{2 \times 85}{9.81}} \approx 4.16 \text{ s}

2. Horizontal velocity:

v_x = 34 \text{ m/s}

3. Final vertical velocity:

v_y = 9.81 \times 4.16 \approx 40.8 \text{ m/s}

Final speed calculation:

v = \sqrt{(34)^2 + (40.8)^2} \approx 53 \text{ m/s}

Thus, the final speed of the soccer ball when it hits the ground is approximately 53 m/s. The answer is (a) 53 m/s.

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- Statistics

Advanced

Proportional Analysis

FRQ

On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 3.5 times as far as he would have on earth, given the same initial velocities on both planets. The ball is launched at a speed of 45 m/s at an angle of 29° above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet find:

- Projectiles

Intermediate

Conceptual

MCQ

Two balls are thrown off a building with the same speed, one straight up and one at a 45° angle. Which statement is true if air resistance can be ignored?

- Energy, Projectiles

Advanced

Mathematical

MCQ

In a lab experiment, a ball is rolled down a ramp so that it leaves the edge of the table with a horizontal velocity v. Assume there are no frictional forces. If the table has a height h above the ground, how far away from the edge of the table, a distance x, does the ball land?

- Projectiles

Advanced

Mathematical

GQ

Suppose the water at the top of Niagara Falls has a horizontal speed of 2.7 m/s just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a 75° angle below the horizontal?

- Projectiles

Advanced

Proportional Analysis

MCQ

- Projectiles

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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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