(a) How far does the textbook travel horizontally after it is released?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_{0x} = v_0 \cos(\theta)\) | Calculate the initial horizontal velocity. Use the initial speed and the angle of projection. |
| 2 | \(v_{0y} = v_0 \sin(\theta)\) | Calculate the initial vertical velocity. Use the initial speed and the angle of projection. |
| 3 | \(v_{0x} = 20 \cos(36^\circ) \approx 16.2 \, \text{m/s}\) | Substitute \( v_0 = 20 \, \text{m/s} \) and \( \theta = 36^\circ \) into the horizontal velocity formula. |
| 4 | \(v_{0y} = 20 \sin(36^\circ) \approx 11.8 \, \text{m/s}\) | Substitute \( v_0 = 20 \, \text{m/s} \) and \( \theta = 36^\circ \) into the vertical velocity formula. |
| 5 | \(y = v_{0y} t – \frac{1}{2} g t^2 + \text{initial height}\) | Use the equation of motion in the vertical direction. The textbook is moving under gravity. |
| 6 | \(0 = 12 + 11.8 t – \frac{1}{2} \cdot 9.8 t^2\) | Set the displacement \( y \) to zero because we are calculating the time \( t \) when the textbook reaches the ground. \( g = 9.8 \text{m/s}^2 \). |
| 7 | \(4.9 t^2 – 11.8 t – 12 = 0\) | Simplify the quadratic equation to solve for \( t \). |
| 8 | \( t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \) | Use the quadratic formula where \( a = 4.9 \), \( b = -11.8 \), and \( c = -12 \). |
| 9 | \( t \approx 3.18 \, \text{s} \) | Solve the equation and take the positive root. This is the time the textbook stays in the air. |
| 10 | \( x = v_{0x} \cdot t \) | Calculate the horizontal distance the textbook travels. Use the horizontal velocity and the time. |
| 11 | \( x \approx 16.2 \times 3.18 \approx 51.5 \, \text{m} \) | Substitute \( v_{0x} = 16.2 \, \text{m/s} \) and \( t = 3.18 \, \text{s} \) into the horizontal distance formula to get the final answer. |
| \( \text{The horizontal distance traveled is approximately } 51.5 \, \text{m}\) | ||
(b) What is the book’s velocity (speed and direction) when it reaches the ground?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_y = v_{0y} – g t \) | Calculate the final vertical velocity using the initial vertical velocity, gravitational acceleration, and time. |
| 2 | \(v_y = 11.8 – 9.8 \times 3.18 \approx -19.4 \, \text{m/s}\) | Substitute \( v_{0y} = 11.8 \, \text{m/s} \), \( g = 9.8 \, \text{m/s}^2 \), and \( t = 3.18 \, \text{s} \) into the vertical velocity formula. |
| 3 | \(v_{\text{total}} = \sqrt{v_{x}^2 + v_y^2} \) | Calculate the magnitude of the total velocity using the Pythagorean theorem. |
| 4 | \(v_{\text{total}} \approx \sqrt{16.2^2 + (-19.4)^2} \approx 25.3 \, \text{m/s} \) | Substitute \( v_{x} = 16.2 \, \text{m/s} \) and \( v_y = -19.4 \, \text{m/s} \) into the total velocity formula. |
| 5 | \(\theta = \tan^{-1} \left(\frac{v_y}{v_x}\right) \) | Calculate the direction of the velocity. Use the inverse tangent to find the angle. |
| 6 | \(\theta \approx \tan^{-1} \left(\frac{-19.4}{16.2}\right) \approx -50.1^\circ \) | The vector points 50.1° below the x-axis. |
| \( \text{The velocity when the book reaches the ground is approximately } 25.3 \, \text{m/s} \text{ at } -50.1^\circ \) | ||
(c) What is the book’s maximum height above the ground?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_y = 0 \) | The vertical velocity at the maximum height is zero. |
| 2 | \(v_y = v_{0y} – g t\) | Use the vertical motion equation to find the time to reach maximum height. |
| 3 | \(0 = 11.8 – 9.8 t\) | Set final vertical velocity \( v_y = 0 \) and solve for \( t \). |
| 4 | \(t = \frac{11.8}{9.8} \approx 1.20 \, \text{s}\) | Solving the equation gives the time to reach maximum height. |
| 5 | \(H = v_{0y} t – \frac{1}{2} g t^2 + \text{initial height}\) | Use the vertical motion equation to find the maximum height. |
| 6 | \(H \approx 11.8 \times 1.2 – \frac{1}{2} \times 9.8 \times (1.2)^2 + 12 \) | Substitute \( v_{0y} = 11.8 \, \text{m/s} \), \( g = 9.8 \, \text{m/s}^2 \), \( t = 1.2 \, \text{s} \), and initial height = 12 m. |
| 7 | \(H \approx 19.1 \, \text{m}\) | Calculate the maximum height above the ground. |
| \( \text{The maximum height above the ground is approximately } 19.1 \, \text{m} \) | ||
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.
Seo-Jun throws a ball to her friend Zuri. The ball leaves Seo-Jun’s hand from a height \( h = 1.5 \) \( \text{m} \) above the ground with an initial speed \( \vec{v}_{s,0} = 12 \) \( \text{m/s} \) at an angle of \( \theta = 25^\circ \) with respect to the horizontal. Zuri catches the ball at a height of \( h = 1.5 \) \( \text{m} \) above the ground.
After catching the ball, Zuri throws it back to Seo-Jun. The ball leaves Zuri’s hand from a height \( h = 1.5 \) \( \text{m} \) above the ground. The ball is moving with a speed of \( 15 \) \( \text{m/s} \) when it reaches a maximum height of \( 5.8 \) \( \text{m} \) above the ground.
At what height \( h’ \) above the ground will the ball be when the return throw reaches Seo-Jun?
Which launch angle gives the greatest horizontal range, assuming level ground and no air resistance?
A ball is launched at an angle. At the peak of its trajectory, which of the following is true?
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?
You must split an apple resting on top of you friend’s head from a distance of 27 m. When you aim directly at the apple, the arrow is horizontal. At what angle should you aim the arrow to hit the apple if the arrow travels at a speed of 35 m/s?
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.
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?
A skier is accelerating down a \( 30.0^{\circ} \) hill at \( 3.80 \) \( \text{m/s}^2 \).
A cat chases a mouse across a \(1.0 \, \text{m}\) high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor \(2.2 \, \text{m}\) from the edge of the table. When the cat slid off the table, what was its speed?
A train is moving to the right at \( 20 \) \( \text{m/s} \). A passenger on the train throws a ball horizontally to the left at \( 5 \) \( \text{m/s} \) (relative to the train).
a) 51.5 m
b) 25.3 m/s at 50.1° below the horizontal
c) 19.1 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
