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- Calculate the time taken for the rocket to reach the wall.
- Calculate the vertical height of the rocket at this time.
- Subtract the height of the wall from this vertical height to find the clearance.

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

1.1 | [katex]v_{x0} = v_0 \cos(\theta)[/katex] | Calculate the horizontal component of the initial velocity. [katex]v_0[/katex] is the launch speed and [katex]\theta[/katex] is the launch angle. |

1.2 | [katex]v_{x0} = 75.0 , \text{m/s} \times \cos(60.0^\circ)[/katex] | Plug in [katex]v_0 = 75.0 , \text{m/s}[/katex] and [katex]\theta = 60.0^\circ[/katex]. |

1.3 | [katex]t = \frac{d}{v_{x0}}[/katex] | Calculate the time [katex]t[/katex] it takes for the rocket to reach the wall. [katex]d[/katex] is the horizontal distance to the wall. |

1.4 | [katex]t = \frac{27.0 , \text{m}}{v_{x0}}[/katex] | Plug in [katex]d = 27.0 , \text{m}[/katex]. |

2.1 | [katex]v_{y0} = v_0 \sin(\theta)[/katex] | Calculate the vertical component of the initial velocity. |

2.2 | [katex]v_{y0} = 75.0 , \text{m/s} \times \sin(60.0^\circ)[/katex] | Plug in [katex]v_0 = 75.0 , \text{m/s}[/katex] and [katex]\theta = 60.0^\circ[/katex]. |

2.3 | [katex]y = v_{y0} t – \frac{1}{2} g t^2[/katex] | Use the kinematic equation for vertical motion. [katex]g[/katex] is the acceleration due to gravity. |

2.4 | [katex]y = v_{y0} t – \frac{1}{2} (9.8 , \text{m/s}^2) t^2[/katex] | Plug in [katex]g = 9.8 , \text{m/s}^2[/katex]. |

2.5 | [katex]y = v_{y0} \times \frac{27.0 , \text{m}}{v_{x0}} – \frac{1}{2} (9.8 , \text{m/s}^2) \left( \frac{27.0 , \text{m}}{v_{x0}} \right)^2[/katex] | Substitute [katex]t[/katex] from step 1.4. |

3.1 | [katex]\text{Clearance} = y – 11.0 , \text{m}[/katex] | Subtract the wall height from the rocket’s height to find the clearance. |

We will now calculate these steps.

The rocket clears the top of the wall by approximately [katex]33.23 , \text{m}[/katex].

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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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Advanced

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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 [katex]v[/katex]. Assume there are no frictional forces. If the table has a height [katex]h[/katex] above the ground, how far away from the edge of the table, a distance [katex]x[/katex], does the ball land?

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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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A circus cannon fires an acrobat into the air at an angle of 45° above the horizontal, and the acrobat reaches a maximum height [katex]y[/katex] above her original launch height. The cannon is now aimed so that it fires straight up into the air at an angle of 90° to the horizontal. In terms of [katex]y[/katex], what is the acrobat’s new max height?

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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?

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33.23 m

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Kinematics | Forces |
---|---|

[katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |

[katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |

[katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |

[katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |

Circular Motion | Energy |
---|---|

[katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |

[katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |

[katex]KE_i + PE_i = KE_f + PE_f[/katex] |

Momentum | Torque and Rotations |
---|---|

[katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |

[katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |

[katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |

Simple Harmonic Motion |
---|

[katex]F = -k x[/katex] |

[katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |

[katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |

Constant | Description |
---|---|

[katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |

[katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |

[katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |

[katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |

[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |

[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |

[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |

Variable | SI Unit |
---|---|

[katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |

[katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |

[katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |

[katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |

[katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |

Variable | Derived SI Unit |
---|---|

[katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |

[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |

[katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |

[katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |

[katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |

[katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |

[katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |

[katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |

General Metric Conversion Chart

Conversion Example

Example of using unit analysis: Convert 5 kilometers to millimeters.

Start with the given measurement:

`[katex]\text{5 km}[/katex]`

Use the conversion factors for kilometers to meters and meters to millimeters:

`[katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]`

Perform the multiplication:

`[katex]\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}[/katex]`

Simplify to get the final answer:

`[katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]`

Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|

Pico- | p | [katex]10^{-12}[/katex] | 0.000000000001 |

Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |

Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |

Milli- | m | [katex]10^{-3}[/katex] | 0.001 |

Centi- | c | [katex]10^{-2}[/katex] | 0.01 |

Deci- | d | [katex]10^{-1}[/katex] | 0.1 |

(Base unit) | – | [katex]10^{0}[/katex] | 1 |

Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |

Hecto- | h | [katex]10^{2}[/katex] | 100 |

Kilo- | k | [katex]10^{3}[/katex] | 1,000 |

Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |

Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |

Tera- | T | [katex]10^{12}[/katex] | 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 [katex] v_i [/katex] is written as [katex] u [/katex]; sometimes [katex] \Delta x [/katex] is written as [katex] s [/katex].
- 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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