Projectile motion range is derived as follows:

- Express the horizontal and vertical components of the initial velocity.
- Determine the time of flight based on vertical motion.
- Calculate the horizontal distance traveled in this time.

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

1.1 | v_{x} = v \cos(\theta) | Horizontal component of initial velocity, constant since no air resistance. |

1.2 | v_{y0} = v \sin(\theta) | Initial vertical component of the velocity. |

2.1 | t = \frac{2v_{y0}}{g} | Time of flight, derived from v_{y} = v_{y0} – gt and the fact that at maximum height, v_{y} = 0, hence t = \frac{2v_{y0}}{g} for a round trip. |

2.2 | t = \frac{2v \sin(\theta)}{g} | Substitute v_{y0} = v \sin(\theta). |

3.1 | R = v_{x} t | Range is the horizontal distance traveled. |

3.2 | R = v \cos(\theta) \times \frac{2v \sin(\theta)}{g} | Substitute v_{x} = v \cos(\theta) and t from step 2.2. |

3.3 | R = \frac{v^2 \sin(2\theta)}{g} | Simplify using trigonometric identity \sin(2\theta) = 2 \sin(\theta) \cos(\theta). |

Now we apply this to find the new range when the launch speed is doubled.

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

4.1 | R' = \frac{(2v)^2 \sin(2\theta)}{g} | New range formula with doubled speed 2v. |

4.2 | R' = 4 \times \frac{v^2 \sin(2\theta)}{g} | Simplify to show the relationship with the original range. |

4.3 | R' = 4R | Substitute the original range R. |

Given the original range of 23 m, we will calculate the new range.

The new range, calculated from the derived formula with the launch speed doubled, is 92.0 , \text{m}.

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- Statistics

Intermediate

Mathematical

FRQ

Measurements made in 1910 indicate that the common flea is an impressive jumper, given its size. Assume that a flea’s initial speed is 2.1 m/s, and that it leaps at an angle of 21° with respect to the horizontal. The jump lasts 0.16 s.

- Projectiles

Intermediate

Conceptual

MCQ

A golfer hits her ball in a high arcing shot. Air resistance is negligible. When the ball is at its highest point, which of the following is true?

- Projectiles

Advanced

Mathematical

FRQ

A block of mass M_{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.

- Momentum, Projectiles

Beginner

Conceptual

MCQ

Which of the following statements about the acceleration due to gravity is TRUE?

- 1D Kinematics, Linear Forces, Projectiles

Advanced

Proportional Analysis

GQ

A ball is kicked at a speed of v_0 an angle \theta above the horizontal. The ball travels 25 meters horizontally. If the ball is kicked at 2v_0 , what will the horizontal displacement be?

- Projectiles

Advanced

Mathematical

FRQ

- Projectiles

Intermediate

Mathematical

GQ

A cat chases a mouse across a 1.0 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 2.2 m from the edge of the table. When the cat slid off the table, what was its speed?

- Projectiles

Advanced

Mathematical

GQ

A diver springs upward from a diving board. At the instant she contacts the water her speed is 8.90 m/s, and her body is extended at an angle of 75.0° with respect to the horizontal surface of the water. At this instant her vertical displacement is -3.00 m, where downward is the negative direction. Determine her initial velocity, both magnitude and direction.

- Projectiles

Advanced

Mathematical

MCQ

A circus cannon fires an acrobat into the air at an angle of 45° above the horizontal, and the acrobat reaches a maximum height y 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 y, what is the acrobat’s new max height?

- 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

92

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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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