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# Part (a): At what times is the object 20 m above the ground?

Step Derivation/Formula Reasoning
1 y = v_0 t + \frac{1}{2} a t^2 Use the kinematic equation for vertical displacement where y is the height, v_0 is the initial velocity, t is the time, and a is the acceleration.
2 20 = 64 t – \frac{1}{2} \cdot 9.8 \cdot t^2 Substitute y = 20 \text{ m}, v_0 = 64 \text{ m/s}, and a = -9.8 \text{ m/s}^2. Note the negative sign of acceleration due to gravity.
3 4.9 t^2 – 64 t + 20 = 0 Rearrange the equation into standard quadratic form a t^2 + b t + c = 0 where a = 4.9, b = -64, and c = 20.
4 t = \frac{64 \pm \sqrt{(-64)^2 – 4 \cdot 4.9 \cdot 20}}{2 \cdot 4.9} Use the quadratic formula t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} to solve for t.
5 t_1 = \frac{64 + 60.88}{9.8} \approx 12.74 \text{ s}, \\ t_2 = \frac{64 – 60.88}{9.8} \approx 0.32 \text{ s} Calculate the two possible values for t. The results are t_1 \approx 12.74 \text{ s} and t_2 \approx 0.32 \text{ s}.
6 t = 0.32 \text{ s}, 12.74 \text{ s} Final answer: There are two times when the object is 20 m above the ground, at 0.32 \text{ s} and 12.74 \text{ s}.

# Part (b): Why are there two answers for part (a)?

Step Explanation Reasoning
1 Two Answers Explanation There are two answers because the object passes 20 m twice: once while going up and once while coming back down.

# Part (c): How long does it take to come back to the ground?

1 y = v_0 t + \frac{1}{2} a t^2 Use the kinematic equation for vertical displacement where y is zero as it returns to the ground.
2 0 = 64 t – \frac{1}{2} \cdot 9.8 \cdot t^2 Substitute y = 0 \text{ m}, v_0 = 64 \text{ m/s}, and a = -9.8 \text{ m/s}^2.
3 0 = 64 t – 4.9 t^2 Simplify the equation by calculating \frac{1}{2} \cdot 9.8 = 4.9.
4 0 = t (64 – 4.9t) Factor out t to solve for the time at which the object returns to the ground.
5 t = 0 \quad \text{or} \quad 64 = 4.9t \Rightarrow t = \frac{64}{4.9} \approx 13.06 \text{ s} The first solution t = 0 corresponds to the initial throw. Solving 64 – 4.9t = 0 gives the time it takes to reach the ground.
6 t = 13.06 \text{ s} Final answer: The object takes 13.06 \text{ s} to come back to the ground.

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1. t = 0.32 \text{ s}, 12.74 \text{ s}
2. There are two answers because the object passes 20 m twice: once while going up and once while coming back down.
3. t = 13.06 \text{ s}

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_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 MotionEnergy
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
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients 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.
kSpring 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
VariableSI 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)}
VariableDerived 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)}
\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.

1. Start with the given measurement: \text{5 km}

2. 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}}

3. 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}

4. 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}

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 v_i is written as u ; sometimes \Delta x is written as s .
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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