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

(a) Calculate the time taken for the ball to reach the wall. |
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1 | Use horizontal motion equation:
\( x = v_0 \cos \theta \cdot t \) |
Relates horizontal distance, initial speed, angle, and time. |

2 | Use vertical motion equation:
\( y = y_0 + v_0 \sin \theta \cdot t – \tfrac{1}{2} g t^2 \) |
Relates vertical position, initial speed, angle, time, and gravity. |

3 | Solve for \( v_0 \) from horizontal equation:
\( v_0 = \dfrac{x}{t \cos \theta} \) |
Express \( v_0 \) in terms of known quantities and \( t \). |

4 | Substitute \( v_0 \) into vertical equation:
\( y = y_0 + \left( \dfrac{x}{t \cos \theta} \right) \sin \theta \cdot t – \tfrac{1}{2} g t^2 \) |
Eliminated \( v_0 \) to get an equation with only \( t \). |

5 | Substitute known values:
\( y = 12.2 \, \text{m} \) \( 12.2 = 1.5 + 45 \times 0.5774 – 4.9 t^2 \) |
Plugged in numerical values to simplify the equation. |

6 | Simplify the equation:
\( 12.2 = 1.5 + 25.983 – 4.9 t^2 \) |
Solved for \( t \), the time to reach the wall. |

(b) Calculate the initial speed of the soccer ball. |
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7 | Use \( v_0 = \dfrac{x}{t \cos \theta} \)
\( \cos 30^\circ = 0.8660 \) |
Calculated initial speed using time from part (a). |

(c) Find the velocity of the soccer ball as it passes over the wall. |
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8 | Compute horizontal velocity component:
\( v_x = v_0 \cos \theta = 29.38 \times 0.8660 \approx 25.45 \, \text{m/s} \) |
Horizontal velocity remains constant. |

9 | Compute vertical velocity at \( t = 1.77 \, \text{s} \):
\( v_y = v_0 \sin \theta – g t \) |
Calculated vertical velocity at the wall (negative indicates downward). |

10 | Compute magnitude of velocity:
\( v = \sqrt{v_x^2 + v_y^2} = \sqrt{(25.45)^2 + (-2.66)^2} \approx 25.58 \, \text{m/s} \) |
Found the speed as it passes over the wall. |

11 | Compute direction angle:
\( \phi = \arctan\left( \dfrac{v_y}{v_x} \right) = \arctan\left( \dfrac{-2.66}{25.45} \right) \approx -5.97^\circ \) |
Determined angle below horizontal. |

(d) Determine if the ball clears the wall when fired at \( 55.5^\circ \). |
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12 | Compute new time to reach wall:
\( \cos 55.5^\circ = 0.5657 \) |
Calculated time to reach wall with new angle. |

13 | Compute vertical position at \( t = 2.72 \, \text{s} \):
\( \sin 55.5^\circ = 0.8286 \) |
Found the height at the wall with new angle. |

14 | Compare height to wall height:
\( y_{\text{wall}} = 9.0 \, \text{m} \) |
Determined that the ball will still clear the wall. |

Just ask: "Help me solve this problem."

- Statistics

Advanced

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FRQ

A ball of mass \( 0.5 \, \text{kg} \), initially at rest, is kicked directly toward a fence from a point \( 32 \, \text{m} \) away, as shown above. The velocity of the ball as it leaves the kicker’s foot is \( 20 \, \text{m/s} \) at an angle of \( 37^\circ \) above the horizontal. The top of the fence is \( 2.5 \, \text{m} \) high. The ball hits nothing while in flight and air resistance is negligible.

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Advanced

Mathematical

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A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.0° above the horizontal. The rocket is fired toward an 11.0-m-high wall, which is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?

- Projectiles

Advanced

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A ball of mass *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*.

- Circular Motion, Projectiles

Intermediate

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A baseball is thrown at an angle of 25° relative to the ground at a speed of 23.0 m/s. The ball is caught 42.0 m from the thrower.

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Advanced

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FRQ

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 [katex] h [/katex] 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 [katex] x [/katex]. The block is released and strikes the floor a horizontal distance [katex] D [/katex] from the edge of the table. Air resistance is negligible.

Derive an expressions for the following quantities only in terms of [katex] M, x, D, h, [/katex] and any constants.

- Energy, Projectiles

(a) The time taken for the ball to reach the wall is approximately \( 1.77 \, \text{s} \).

(b) The initial speed of the soccer ball is approximately \( 29.38 \, \text{m/s} \).

(c) As it passes over the wall, the ball’s velocity is approximately \( 25.58 \, \text{m/s} \) at an angle of \( 5.97^\circ \) below the horizontal.

(d) Yes, when fired at \( 55.5^\circ \), the ball will still clear the \( 9.0 \, \text{m} \) high wall, passing over it at a height of approximately \( 31.64 \, \text{m} \).

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

\(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |

\(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |

\(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |

\(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |

\(v^2 = v_f^2 \,-\, 2a \Delta x\) |

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

\(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |

\(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |

\(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |

\(W = Fd \cos\theta\) |

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

\(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |

\(J = \Delta p\) | \(I = \sum mr^2\) |

\(p_i = p_f\) | \(L = I \cdot \omega\) |

Simple Harmonic Motion | Fluids |
---|---|

\(F = -kx\) | \(P = \frac{F}{A}\) |

\(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |

\(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |

\(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |

\(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |

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 |

- 1. Some answers may vary by 1% due to rounding.
- Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
- Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
- Bookmark questions you can’t solve to revisit them later
- 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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