You kick a soccer ball with an initial velocity directed 53° above the horizontal. The ball lands on a roof 7.2 m high. The wall of the building is 25 m away, and it takes the ball 2.1 seconds to pass directly over the wall.

- (a) Calculate the initial velocity of the ball.
*(2 points)* - (b) Determine the horizontal range of the ball.
*(2 points)* - (c) By what vertical distance does the ball clear the wall of the building?
*(3 points)*

- 19.8 m/s
- 32.2 m
- 5.52 m

First, calculate the initial velocity of the ball.

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

1 | y = v_{0y} t – \frac{1}{2} g t^2 | Vertical motion equation for height (y), initial vertical velocity (v_{0y}), time (t), and acceleration due to gravity (g). |

2 | v_{0y} = \frac{y + \frac{1}{2} g t^2}{t} | Solve for initial vertical velocity (v_{0y}). |

3 | v_{0x} = \frac{d}{t} | Horizontal velocity (v_{0x}) is constant, where d is the distance to the wall. |

4 | v_{0} = \sqrt{v_{0x}^2 + v_{0y}^2} | Initial velocity magnitude using Pythagorean theorem, combining horizontal and vertical components. |

Given values:

- Wall height y = 7.2 , \text{m}
- Distance to wall d = 25 , \text{m}
- Time to pass over wall t = 2.1 , \text{s}
- Acceleration due to gravity g = -9.81 , \text{m/s}^2

Next, determine the horizontal range of the ball.

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

1 | R = v_{0x} T | Horizontal range (R), where T is the total time of flight. |

2 | \Delta y = v_{oy}t \frac{1}{2}gt^2 | Total time of flight from launch to landing on rood, using symmetry of projectile motion. T = 2.65 seconds. |

Finally, calculate the vertical distance the ball clears the wall.

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

1 | h_{clear} = y_{peak} – y_{wall} | Vertical clearance (h_{clear}) is the difference between the peak height (y_{peak}) and wall height (y_{wall}). |

2 | y_{peak} = \frac{v_{0y}^2}{2g} | Peak height calculation using the initial vertical velocity. Peak height = 12.7 m. |

The calculations yield the following results:

- Initial velocity of the ball: \boxed{19.8 , \text{m/s}}
- Horizontal range of the ball: \boxed{32.2 , \text{m}}
- Vertical distance by which the ball clears the wall: \boxed{5.52 , \text{m}}

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The highest barrier that a projectile can clear is 16.2 m, when the projectile is launched at an angle of 22.0° above the horizontal. What is the projectile’s launch speed?

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MCQ

A plane, 220 meters high, is dropping a supply crate to an island below. It is traveling with a horizontal velocity of 150 m/s. At what horizontal distance must the plane drop the supply crate for it to land on the island?

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*t _{A}*,

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Proportional Analysis

MCQ

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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?

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Advanced

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In the absence of air resistance, a projectile is launched from, returns to ground level and has a range of 23 m. Suppose the launch speed is doubled, and the projectile is fired at the same angle above the ground. What is the new range?

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GQ

A person shoots a basket ball with a speed of 12 m/s at an angle of 35° above the horizontal. If the person is 2.4 m tall and the hoop is 3.05 m above the ground, how far back must the person stand in order to make the shot?

- Projectiles

Intermediate

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MCQ

A rock is thrown at an angle of 42° above the horizontal at a speed of 14 m/s. Determine how long it takes the rock to hit the ground.

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Intermediate

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GQ

A golfer hits a shot to a green that is elevated 2.80 m above the point where the ball is struck. The ball leaves the club at a speed of 18.9 m/s at an angle of 52.0° above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.

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FRQ

A javelin thrower standing at rest holds the center of the javelin behind her head, then accelerates it through a distance of 70 cm as she throws. She releases the 600 g javelin 2.0 m above the ground traveling at an angle of 30° above the horizontal. In this throw, the javelin hits the ground 54 m away. Find the following:

- Projectiles

- 19.8 m/s
- 32.2 m
- 5.52 m

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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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