| Derivation/Formula | Reasoning |
|---|---|
| \[a_y = a \sin \theta\] | The vertical component \(a_y\) equals the downhill acceleration \(a\) multiplied by \(\sin \theta\). |
| \[a_y = (3.80\,\text{m/s}^2)\sin 30^\circ\] | Insert \(a = 3.80\,\text{m/s}^2\) and \(\theta = 30^\circ\). |
| \[a_y = 1.90\,\text{m/s}^2\] | Since \(\sin 30^\circ = 0.5\), multiplying yields \(1.90\,\text{m/s}^2\). |
| \[\boxed{a_y = 1.90\,\text{m/s}^2 \text{ downward}}\] | Vertical component magnitude and direction. |
| Derivation/Formula | Reasoning |
|---|---|
| \[s = \frac{\Delta y}{\sin \theta}\] | The path length \(s\) relates to vertical drop \(\Delta y\) via \(\sin \theta\). |
| \[s = \frac{250\,\text{m}}{\sin 30^\circ}\] | Substitute \(\Delta y = 250\,\text{m}\) and \(\theta = 30^\circ\). |
| \[s = 500\,\text{m}\] | Compute \(s\) using \(\sin 30^\circ = 0.5\). |
| \[s = \frac{1}{2} a t^2\] | With zero initial speed, displacement along slope is \(s = \frac{1}{2} a t^2\). |
| \[t = \sqrt{\frac{2s}{a}}\] | Solve for \(t\). |
| \[t = \sqrt{\frac{2(500\,\text{m})}{3.80\,\text{m/s}^2}}\] | Substitute \(s = 500\,\text{m}\) and \(a = 3.80\,\text{m/s}^2\). |
| \[\boxed{t = 16.2\,\text{s}}\] | Time to reach the bottom. |
A Major Upgrade To Phy Is Coming Soon — Stay Tuned
We'll help clarify entire units in one hour or less — guaranteed.
A self paced course with videos, problems sets, and everything you need to get a 5. Trusted by over 15k students and over 200 schools.
Two cannonballs, A and B, are fired from the ground with identical initial speeds, but with \( \theta_A \) larger than \( \theta_B \).
A officer fires a pistol horizontally toward a target \(120 \,\text{m}\) at a velocity of \(200 \, \text{m/s}\). If the officer aimed directly at the bull’s eye
A person is standing at the edge of the water and looking out at the ocean. The height of the person’s eyes above the water is \( h = 1.8 \, \text{m} \), and the radius of the Earth is \( R = 6.38 \times 10^6 \, \text{m} \). How far is it to the horizon (in meters)? In other words, find the distance \( d \) from the person’s eyes to the horizon. Note at the horizon, the angle between the line of sight and the radius of the Earth is \( 90^\circ \).)
A student walks \( 3 \) \( \text{m} \) east, then \( 4 \) \( \text{m} \) west in \( 7 \) \( \text{s} \). What is their displacement and average velocity?
What does displacement mean in the context of motion?
A person shoots a basketball with a speed of \( 12 \, \text{m/s} \) at an angle of \( 35^\circ \) above the horizontal. If the person is \( 2.4 \, \text{m} \) tall and the hoop is \( 3.05 \, \text{m} \) above the ground, how far back must the person stand in order to make the shot?
A bald eagle in level flight at a height of \(135 \, \text{m}\) drops the fish it caught. If the eagle’s speed is \(25.0 \, \text{m/s}\) how far from the drop point will the fish land?
In the absence of air resistance, a projectile is launched from and returns to ground level and has a range of \( 23 \, \text{m} \). Suppose the launch speed is doubled, and the projectile is fired at the same angle above the ground. What is the new range?
3 clay balls, labeled A, B, and C are launched from the same height at the same speed as shown above. A is launched at \( 30^\circ \) above horizontal, B is launched horizontally, and C is launched \( 30^\circ \) below the horizontal. They all hit the wall (before reaching the ground) in times \( t_A \), \( t_B \), and \( t_C \) respectively. Rank these times from least to greatest.
An eagle is flying horizontally at \(6 \, \text{m/s}\) with a fish in its claws. It accidentally drops the fish.
By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
Metric Prefixes
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] |
One price to unlock most advanced version of Phy across all our tools.
per month
Billed Monthly. Cancel Anytime.
We crafted THE Ultimate A.P Physics 1 Program so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the 2026 AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed
