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# Part (a): What is the difference in the time the balls spend in the air?

Step Derivation/Formula Reasoning
1 y = v_0 t + \frac{1}{2}gt^2 Use the kinematic equation for vertical motion to find the time of flight of each ball.
2 19.6 = 0 + 14.7t + \frac{1}{2}(9.8)t^2 For the ball thrown downward: [katex] v_0 = -14.7 \, \text{m/s} [/katex], [katex] g = 9.8 \, \text{m/s}^2 [/katex]
3 19.6 = 14.7t + 4.9t^2 \Rightarrow 0 = 4.9t^2 + 14.7t – 19.6 Solve the quadratic equation using the quadratic formula.
4 t = \frac{-14.7 \pm \sqrt{(14.7)^2 – 4(4.9)(-19.6)}}{2(4.9)} The positive root is taken since time cannot be negative.
5 t = 1 \, \text{s} Solving the quadratic equation gives [katex] t = 1 \, \text{s} [/katex].
6 y = v_0 t + \frac{1}{2}gt^2 Now for the ball thrown upward: initial speed [katex] v_0 = 14.7 \, \text{m/s}
7 0 = 14.7t – \frac{1}{2}(9.8)t^2 \Rightarrow t(2.9 – 2.9t) = 0 \Rightarrow t = 3 \, \text{s} Solving for \( t [/katex] when the ball reaches the top of its trajectory.
8 \text{Total time in air} = (2 \times 1 + 3)\ s = 5 \, \text{.} The time the second ball spends in the air is double the ascent time plus the fall time.
9 Difference in time = 2 \, \text{s} – 1\,\text{s} = 1s The difference in the time the balls spend in the air.

# Part (b): What is the velocity of each ball as it strikes the ground?

Step Derivation/Formula Reasoning
1 v = v_0 + gt Use the first kinematic equation to find the final velocity.
2 v = -14.7 + 9.8 \times 1 = -24.5 \, \text{m/s} For the ball thrown downward: add gravitational acceleration effect.
3 v = 14.7 – 9.8 \times 3 = -14.7 \, \text{m/s} For the ball thrown upward: subtract the effect of gravity over fall time.

# Part (c): How far apart are the balls 0.800 s after they are thrown?

Step Derivation/Formula Reasoning
1 d_{downward} = v_0 t + \frac{1}{2}gt^2 Use kinematic equation for the ball thrown downward.
2 d_{downward} = 14.7 \times 0.800 + \frac{1}{2} \times 9.8 \times 0.800^2 = 15.68 \, \text{m} Calculate distance traveled by the downward-thrown ball in 0.800s.
3 d_{upward} = v_0 t – \frac{1}{2}gt^2 Use kinematic equation for the ball thrown upward.
4 d_{upward} = 14.7 \times 0.800 – \frac{1}{2} \times 9.8 \times 0.800^2 = 6.864 \, \text{m} Calculate distance traveled by the upward-thrown ball in 0.800s.
5 20 \, \text{m} Distance between the balls after 0.800 s is [katex]15.68\, \text{m} + 6.864\, \text{m}[/katex] = 22.544

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1. Difference in Time Spent in the Air: [katex]\boxed{\Delta t = 0.5 , \text{seconds}}[/katex]
2. Velocity of Each Ball as it Strikes the Ground:
• Ball Thrown Downward: [katex]\boxed{v_{\text{down}} = 53.9 , \text{m/s}}[/katex]
• Ball Thrown Upward: [katex]\boxed{v_{\text{up}} = 19.6 , \text{m/s}}[/katex]
3. Distance Apart after 0.800 Seconds: [katex]\boxed{\text{Distance Apart} = 2.352 , \text{meters}}[/katex]

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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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