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(a) How far does the textbook travel horizontally after it is released?

Step Derivation/Formula Reasoning
1 [katex]v_{0x} = v_0 \cos(\theta)[/katex] Calculate the initial horizontal velocity. Use the initial speed and the angle of projection.
2 [katex]v_{0y} = v_0 \sin(\theta)[/katex] Calculate the initial vertical velocity. Use the initial speed and the angle of projection.
3 [katex]v_{0x} = 20 \cos(36^\circ) \approx 16.2 \, \text{m/s}[/katex] Substitute [katex] v_0 = 20 \, \text{m/s} [/katex] and [katex] \theta = 36^\circ [/katex] into the horizontal velocity formula.
4 [katex]v_{0y} = 20 \sin(36^\circ) \approx 11.8 \, \text{m/s}[/katex] Substitute [katex] v_0 = 20 \, \text{m/s} [/katex] and [katex] \theta = 36^\circ [/katex] into the vertical velocity formula.
5 [katex]y = v_{0y} t – \frac{1}{2} g t^2 + \text{initial height}[/katex] Use the equation of motion in the vertical direction. The textbook is moving under gravity.
6 [katex]0 = 12 + 11.8 t – \frac{1}{2} \cdot 9.8 t^2[/katex] Set the displacement [katex] y [/katex] to zero because we are calculating the time [katex] t [/katex] when the textbook reaches the ground. [katex] g = 9.8 \text{m/s}^2 [/katex].
7 [katex]4.9 t^2 – 11.8 t – 12 = 0[/katex] Simplify the quadratic equation to solve for [katex] t [/katex].
8 [katex] t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} [/katex] Use the quadratic formula where [katex] a = 4.9 [/katex], [katex] b = -11.8 [/katex], and [katex] c = -12 [/katex].
9 [katex] t \approx 3.18 \, \text{s} [/katex] Solve the equation and take the positive root. This is the time the textbook stays in the air.
10 [katex] x = v_{0x} \cdot t [/katex] Calculate the horizontal distance the textbook travels. Use the horizontal velocity and the time.
11 [katex] x \approx 16.2 \times 3.18 \approx 51.5 \, \text{m} [/katex] Substitute [katex] v_{0x} = 16.2 \, \text{m/s} [/katex] and [katex] t = 3.18 \, \text{s} [/katex] into the horizontal distance formula to get the final answer.
[katex] \text{The horizontal distance traveled is approximately } 51.5 \, \text{m}[/katex]

(b) What is the book’s velocity (speed and direction) when it reaches the ground?

Step Derivation/Formula Reasoning
1 [katex]v_y = v_{0y} – g t [/katex] Calculate the final vertical velocity using the initial vertical velocity, gravitational acceleration, and time.
2 [katex]v_y = 11.8 – 9.8 \times 3.18 \approx -19.4 \, \text{m/s}[/katex] Substitute [katex] v_{0y} = 11.8 \, \text{m/s} [/katex], [katex] g = 9.8 \, \text{m/s}^2 [/katex], and [katex] t = 3.18 \, \text{s} [/katex] into the vertical velocity formula.
3 [katex]v_{\text{total}} = \sqrt{v_{x}^2 + v_y^2} [/katex] Calculate the magnitude of the total velocity using the Pythagorean theorem.
4 [katex]v_{\text{total}} \approx \sqrt{16.2^2 + (-19.4)^2} \approx 25.3 \, \text{m/s} [/katex] Substitute [katex] v_{x} = 16.2 \, \text{m/s} [/katex] and [katex] v_y = -19.4 \, \text{m/s} [/katex] into the total velocity formula.
5 [katex]\theta = \tan^{-1} \left(\frac{v_y}{v_x}\right) [/katex] Calculate the direction of the velocity. Use the inverse tangent to find the angle.
6 [katex]\theta \approx \tan^{-1} \left(\frac{-19.4}{16.2}\right) \approx -50.1^\circ [/katex] The vector points 50.1° below the x-axis.
[katex] \text{The velocity when the book reaches the ground is approximately } 25.3 \, \text{m/s} \text{ at } -50.1^\circ [/katex]

(c) What is the book’s maximum height above the ground?

Step Derivation/Formula Reasoning
1 [katex]v_y = 0 [/katex] The vertical velocity at the maximum height is zero.
2 [katex]v_y = v_{0y} – g t[/katex] Use the vertical motion equation to find the time to reach maximum height.
3 [katex]0 = 11.8 – 9.8 t[/katex] Set final vertical velocity [katex] v_y = 0 [/katex] and solve for [katex] t [/katex].
4 [katex]t = \frac{11.8}{9.8} \approx 1.20 \, \text{s}[/katex] Solving the equation gives the time to reach maximum height.
5 [katex]H = v_{0y} t – \frac{1}{2} g t^2 + \text{initial height}[/katex] Use the vertical motion equation to find the maximum height.
6 [katex]H \approx 11.8 \times 1.2 – \frac{1}{2} \times 9.8 \times (1.2)^2 + 12 [/katex] Substitute [katex] v_{0y} = 11.8 \, \text{m/s} [/katex], [katex] g = 9.8 \, \text{m/s}^2 [/katex], [katex] t = 1.2 \, \text{s} [/katex], and initial height = 12 m.
7 [katex]H \approx 19.1 \, \text{m}[/katex] Calculate the maximum height above the ground.
[katex] \text{The maximum height above the ground is approximately } 19.1 \, \text{m} [/katex]

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a) 51.5 m

b) 25.3 m/s at 50.1° below the horizontal

c) 19.1 m

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KinematicsForces
[katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex][katex]F = ma[/katex]
[katex]v = v_i + at[/katex][katex]F_g = \frac{G m_1m_2}{r^2}[/katex]
[katex]a = \frac{\Delta v}{\Delta t}[/katex][katex]f = \mu N[/katex]
[katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex]
Circular MotionEnergy
[katex]F_c = \frac{mv^2}{r}[/katex][katex]KE = \frac{1}{2} mv^2[/katex]
[katex]a_c = \frac{v^2}{r}[/katex][katex]PE = mgh[/katex]
[katex]KE_i + PE_i = KE_f + PE_f[/katex]
MomentumTorque and Rotations
[katex]p = m v[/katex][katex]\tau = r \cdot F \cdot \sin(\theta)[/katex]
[katex]J = \Delta p[/katex][katex]I = \sum mr^2[/katex]
[katex]p_i = p_f[/katex][katex]L = I \cdot \omega[/katex]
Simple Harmonic Motion
[katex]F = -k x[/katex]
[katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex]
[katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex]
ConstantDescription
[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
VariableSI 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]
VariableDerived 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]\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]

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: [katex]\text{5 km}[/katex]

2. 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]

3. 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]

4. 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]

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity [katex] v_i [/katex] is written as [katex] u [/katex]; sometimes [katex] \Delta x [/katex] is written as [katex] s [/katex].
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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