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Part a:

Step Derivation/Formula Reasoning
1 y = y_0 + v_0 t – \frac{1}{2} g t^2 This is the kinematic equation for vertical position, where y_0 is the initial position, v_0 is the initial velocity, g is the acceleration due to gravity, and t is the time.
2 Substitute y = 0, y_0 = 800\ \text{m}, v_0 = 65\ \text{m/s}, g = 9.8\ \text{m/s}^2 The object hits the ground when y=0. It is thrown upward from a height of 800 m and the initial velocity is upward, hence positive.
3 0 = 800 + 65t – 4.9t^2 Rearrange the substituted equation and simplify g/2 from 9.8/2 to 4.9.
4 4.9t^2 – 65t – 800 = 0 This is a quadratic equation in the form of at^2 + bt + c = 0.
5 t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} Apply the quadratic formula to solve for t. Here, a = 4.9, b = -65, and c = -800.
6 t = \frac{-(-65) \pm \sqrt{(-65)^2 – 4 \cdot 4.9 \cdot (-800)}}{2 \cdot 4.9} Plug in the values of a, b, and c.
10 t \approx 21.03\ \text{s} \textbf{ (positive root)} Discard negative time and keep the physically meaningful positive root, being the actual time the object takes to hit the ground.

Part b:

Step Derivation/Formula Reasoning
1 v = v_0 – gt Use the kinematic equation for velocity in vertical motion, taking downward as the negative direction.
2 v = 65 – 9.8 \times 21.04 Substitute the values of g and t into the equation.
3 v \approx -141.14\ \text{m/s} Subtract to find the final velocity, where negative indicates direction downwards.

Part c:

Step Derivation/Formula Reasoning
1 v \propto v_0 The final velocity is proportional to the initial velocity when all other factors remain the same.
2 v_{\text{final, new}} = 2v_0 – gt Note this is the equation we used in part b to find the final speed. Time, however, is not the same in the two scenarios. So unless you have found time with the double the initial speed, do NOT use this equation again.
3 v_{\text{final, new}} = {2v_0}^2 + 2g\Delta \theta Use this kinematic equation instead.
4 {v_{\text{final, new}}}^2 = = 180.54 \, m/s Plug in values and solve for the new final speed.
5 \frac{v_{\text{final, new}}}{v_{\text{final, orginal}} Find the factor the final velocity has increased by.
6 \frac{180.54}{141.14} \approx 1.28 Plug in values and solve.
7 Factor = 1.28 If the initial velocity doubles the final velocity of the ball increases by a factor of 1.28.

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1. t \approx 21.03\ \text{s}
2. v \approx 141.14\ \text{m/s} downwards.
3. The final speed is directly proportional to the initial velocity and will increase by \approx 1.28 times. This can be proven by any one of the kinematic equations such as  as proven by the kinematic equation {v_f}^2 = {v_0}^2 + 2a\Delta \theta.

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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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