(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} | ||
Phy can also check your working. Just snap a picture!
A baseball rolls off a 0.70 m high desk and strikes the floor 0.25 m away from the base of the desk. How fast was the ball rolling?
A golfer hits a shot to a green that is elevated 2.80 m above the point where the ball is struck. The ball leaves the club at a speed of 18.9 m/s at an angle of 52.0° above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
A ball is kicked horizontally off a 20 m tall cliff at a speed of 11 m/s. What is the final velocity of the ball right before it hits the ground?
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 h 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 x . The block is released and strikes the floor a horizontal distance D from the edge of the table. Air resistance is negligible.
Derive an expressions for the following quantities only in terms of M, x, D, h, and any constants.
Measurements made in 1910 indicate that the common flea is an impressive jumper, given its size. Assume that a flea’s initial speed is 2.1 m/s, and that it leaps at an angle of 21° with respect to the horizontal. The jump lasts 0.16 s.
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 Sale and Terms of Use 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_1m_2}{r^2} |
| a = \frac{\Delta v}{\Delta t} | f = \mu N |
| R = \frac{v_i^2 \sin(2\theta)}{g} |
| Circular Motion | Energy |
|---|---|
| F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
| a_c = \frac{v^2}{r} | PE = mgh |
| KE_i + PE_i = KE_f + PE_f |
| Momentum | Torque and Rotations |
|---|---|
| p = m v | \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 |
|---|
| F = -k x |
| T = 2\pi \sqrt{\frac{l}{g}} |
| T = 2\pi \sqrt{\frac{m}{k}} |
| Constant | Description |
|---|---|
| g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
| G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
| \mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
| k | Spring constant, in \text{N/m} |
| M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
| M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
| M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| s (Displacement) | \text{meters (m)} |
| v (Velocity) | \text{meters per second (m/s)} |
| a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
| t (Time) | \text{seconds (s)} |
| m (Mass) | \text{kilograms (kg)} |
| Variable | Derived SI Unit |
|---|---|
| F (Force) | \text{newtons (N)} |
| E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
| P (Power) | \text{watts (W)} |
| p (Momentum) | \text{kilogram meters per second (kgm/s)} |
| \omega (Angular Velocity) | \text{radians per second (rad/s)} |
| \tau (Torque) | \text{newton meters (Nm)} |
| I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |
| f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \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}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | 10^{-12} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
The most advanced version of Phy. Currently 50% off, for early supporters.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Understand you mistakes quicker.
For GQs and FRQs, Phy provides brief feedback as to how you can improve your answer.
Aim to increase your understadning and average score with every attempt!
10 Free Credits To Get You Started
*Phy Pro members get unlimited credits
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
