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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v_f = v_i + at[/katex] | Use the kinematic equation to find the time to reach the maximum height. Here, [katex]v_f[/katex] is the final velocity (0 m/s), [katex]v_i[/katex] is the initial velocity (25.3 m/s), [katex]a[/katex] is the acceleration due to gravity (-9.8 m/s²), and [katex]t[/katex] is the time. |
| 2 | [katex]0 = 25.3 \, \text{m/s} – (9.8 \, \text{m/s}^2) t[/katex] | Substitute the known values into the kinematic equation. |
| 3 | [katex]t = \frac{25.3 \, \text{m/s}}{9.8 \, \text{m/s}^2}[/katex] | Solve for [katex]t[/katex], the time to reach maximum height. |
| 4 | [katex]t \approx 2.58 \, \text{s}[/katex] | Calculate the time numerically. |
| 5 | [katex]y_f = v_i t + \frac{1}{2}at^2[/katex] | Use the kinematic equation to find the total displacement from the initial point to the roof after maximum height. Here, [katex]y_f[/katex] is the final height (17.4 m), [katex]a[/katex] is the acceleration due to gravity (-9.8 m/s²), and [katex]v_i[/katex] is 0 since it starts falling from rest at maximum height. |
| 6 | [katex]17.4 = 0 + \frac{1}{2}(-9.8)t^2[/katex] | Substitute the values into the kinematic equation. We need to solve for the time [katex]t[/katex] it takes to fall back down 17.4 m. |
| 7 | [katex]17.4 = -4.9t^2[/katex] | Simplify the equation by calculating the constant in the kinematic equation. |
| 8 | [katex]t^2 = \frac{17.4}{4.9}[/katex] | Rearrange the equation to isolate [katex]t^2[/katex] on one side of the equation. |
| 9 | [katex]t = \sqrt{\frac{17.4}{4.9}}[/katex] | Solve for [katex]t[/katex]. |
| 10 | [katex]t \approx 1.88 \, \text{s}[/katex] | Calculate the time numerically for the descent from the maximum height to the roof level. |
| 11 | [katex] \mathbf{t_{\text{total}} = t_{\text{up}} + t_{\text{down}} = 2.58 \text{s} + 1.88 \text{s}} [/katex] | Add the time to reach the maximum height and the time to descend to the roof level. |
| 12 | [katex]\boxed{4.46 \text{s}}[/katex] | Final answer: the total time taken for the baseball to reach the roof level. |
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You stand at the edge of a vertical cliff and throws a stone vertically upwards. The stone leaves your hand with a speed v = 8.0 m/s. The time between the stone leaving your hand and hitting the sea is 3.0 s. Assume air resistance is negligible. Calculate:
The first \(10 \, \text{meters}\) of a \(100 \, \text{meter}\) dash are covered in \(2 \, \text{seconds}\) by a sprinter who starts from rest and accelerates with a constant acceleration. The remaining \(90 \, \text{meters}\) are run with the same velocity the sprinter had after \(2 \, \text{seconds}\).
A ranger in a national park is driving at 56 km/h when a deer jumps onto the road 65 m ahead of the vehicle. After a reaction time of t s, the ranger applies the brakes to produce an acceleration of -3 m/s2. What is the maximum reaction time allowed if the ranger is to avoid hitting the deer?
Consider a ball thrown up from the surface of the earth into the air at an angle of 30° above the horizontal. Air resistance is negligible. The ball’s acceleration just after release is most nearly
A red car, initially at rest, travels east with an acceleration of 3.5 m/s2. At the same time as the red car starts to move, a blue car is traveling west at 15 m/s and accelerating at 1.2 m/s2. If they are 600 m apart the moment the red car starts to move and they are traveling towards each other, where and when will they meet?
4.46 seconds
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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
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] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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