0 attempts
0% avg
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(\Delta y = 80 \, \text{m}\) | Identify the vertical displacement from the initial height of the cliff to the ground. |
| 2 | \(\Delta x = ?\) | Determine the horizontal displacement from the base of the cliff. |
| 3 | \(v_i = 8.0 \, \text{m/s}\) | Identify the initial horizontal velocity of the stone. |
| 4 | \(v_{y0} = 0 \, \text{m/s}\) | The initial vertical velocity of the stone is zero because it is thrown horizontally. |
| 5 | \(\Delta y = v_{y0} t + \frac{1}{2} g t^2\) | Use the equation of motion vertically to determine the time of flight. Here, \( \Delta y \) is the vertical displacement, \( v_{y0} \) is the initial vertical velocity, \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)), and \( t \) is the time. |
| 6 | \(80 = 0 + \frac{1}{2} \cdot 9.8 \cdot t^2\) | Substitute the known values into the vertical motion equation. |
| 7 | \(80 = 4.9 \cdot t^2\) | Simplify the equation. |
| 8 | \(t^2 = \frac{80}{4.9}\) | Rearrange the equation to solve for \( t^2 \). |
| 9 | \(t^2 \approx 16.33\) | Calculate the value on the right-hand side. |
| 10 | \(t \approx \sqrt{16.33}\) | Take the square root of both sides to solve for \( t \). |
| 11 | \(t \approx 4.04 \, \text{s}\) | Calculate the time of flight. |
| 12 | \(\Delta x = v_i t\) | Use the equation of motion horizontally to determine the horizontal displacement. Here, \( v_i \) is the initial horizontal velocity. |
| 13 | \(\Delta x = 8.0 \cdot 4.04\) | Substitute the known values into the horizontal motion equation. |
| 14 | \(\Delta x \approx 32.32 \, \text{m}\) | Calculate the horizontal displacement. |
| 15 | \(\boxed{\Delta x \approx 32.32 \, \text{m}}\) | State the final result, boxed: The stone will strike the ground approximately 32.32 meters from the base of the cliff. |
Just ask: "Help me solve this problem."
We'll help clarify entire units in one hour or less — guaranteed.
A major-league pitcher can throw a baseball in excess of \( 41.0 \, \text{m/s} \). If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is \( 17.0 \, \text{m} \) away from the point of release?
A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge that is \( 235 \) \( \text{m} \) below. The plane is traveling horizontally with a speed of \( 250 \) \( \text{km/h} \). How far in advance of the recipients (horizontal distance) must the goods be dropped?
A projectile is launched at angle \( \theta \) to the horizontal, with velocity \( v \), maximum vertical displacement \( s \), and angle \( \theta \) between \( 0^{\circ} \) and \( 45^{\circ} \). What will the maximum vertical displacement be if the projectile is now launched at an angle of \( 2 \theta \) from the horizontal with velocity \( v \)?
A soccer ball is kicked horizontally off an \( 85 \) \( \text{m} \) high cliff at a speed of \( 34 \) \( \text{m/s} \). What is the ball’s final speed when it hits the ground below?
A projectile is launched at an upward angle of \( 30^\circ \) to the horizontal with a speed of \( 30 \) \( \text{m/s} \). How does the horizontal component of its velocity \( 1.0 \) \( \text{s} \) after launch compare with its horizontal component of velocity \( 2.0 \) \( \text{s} \) after launch, ignoring air resistance?
An airplane with a speed of \( 97.5 \, \text{m/s} \) is climbing upward at an angle of \( 50.0^\circ \) with respect to the horizontal. When the plane’s altitude is \( 732 \, \text{m} \), the pilot releases a package.
Which of the following statements about the acceleration due to gravity is TRUE?
A projectile has the least speed at what point in its path?
Person A throws a ball horizontally from a cliff \( 20 \) \( \text{m} \) tall at \( 12 \) \( \text{m/s} \). Person B is running to the right on the ground and catches the ball at the same height it would’ve landed after running \( 15 \) \( \text{m} \). How fast was Person B running?
A bird, traveling at \(50 \, \text{m/s}\) wants to hit a man \(100 \, \text{m}\) below with a dropping. How far in distance before flying directly over the man should the bird release it?
32 m
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
