(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?
A car accelerates from rest with an acceleration of \( 3.5 \, \text{m/s}^2 \) for \( 10 \, \text{s} \). After this, it continues at a constant speed for an unknown amount of time. The driver notices a ramp \( 50 \, \text{m} \) ahead and takes \( 0.6 \, \text{s} \) to react. After reacting, the driver hits the brakes, which slow the car with an acceleration of \( 7.2 \, \text{m/s}^2 \). Unfortunately, the driver does not stop in time and goes off the \( 3 \, \text{m} \) high ramp that is angled at \( 27^\circ \).
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?
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.
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?
An eagle is flying horizontally at \(6 \, \text{m/s}\) with a fish in its claws. It accidentally drops the fish.
A block of mass \(M_1\) travels horizontally with a constant speed \(v_0\) on a plateau of height \(H\) until it comes to a cliff. A toboggan of mass \(M_2\) is positioned on level ground below the cliff. The center of the toboggan is a distance \(D\) from the base of the cliff.
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?
In archery, should the arrow be aimed directly at the target? How should your angle of aim depend on the distance to the target? Explain without using equations.
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) 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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