(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. |
| \( \text{The maximum height above the ground is approximately } 19.1 \, \text{m} \) | ||
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A ball is launched at an angle. At the peak of its trajectory, which of the following is true?
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 expressions for the following quantities only in terms of \(M, x, D, h,\) and any constants.
The highest barrier that a projectile can clear is 16.2 m, when the projectile is launched at an angle of 22.0° above the horizontal. What is the projectile’s launch speed?
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?
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?
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?
Suppose the water at the top of Niagara Falls has a horizontal speed of \( 2.7 \, \text{m/s} \) just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a \( 75^\circ \) angle below the horizontal?
A soccer ball with an initial height of \(1.5 \, \text{m}\) above the ground is launched at an angle of \(30^\circ\) above the horizontal. The soccer ball travels a horizontal distance of \(45 \, \text{m}\) to a \(9.0 \, \text{m}\) high castle wall, and passes over \(3.20 \, \text{m}\) above the highest point of the wall. Assume air resistance is negligible.
A ball is launched and lands \( 20 \) \( \text{m} \) away below the launch point \( 2.5 \) \( \text{s} \) later. The maximum height reached is \( 8.0 \) \( \text{m} \). What was the original launch velocity?
A ball is shot from the top of a building with an initial velocity of \( 18 \) \( \text{m/s} \) at an angle \( \theta = 42^\circ \) above the horizontal.
a) 51.5 m
b) 25.3 m/s at 50.1° below the horizontal
c) 19.1 m
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| 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] |
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