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# Part (a): Calculate the time taken for the ball to reach the wall.

Step Derivation/Formula Reasoning
1 x = v_0 \cos(\theta) t Using the horizontal motion formula where x is the horizontal displacement, v_0 is the initial velocity, \theta is the angle of launch, and t is time.
2 t = \frac{x}{v_0 \cos(\theta)} Isolating t to solve for the time taken to reach the wall.
3 x = 45 \, \text{m}, \theta = 30^\circ Substitute the given values of x and \theta into the equation. Note that we need v_0, which will be found in part (b).

# Part (b): Calculate the initial speed of the soccer ball.

Step Derivation/Formula Reasoning
1 y = y_0 + v_0 \sin(\theta) t – \frac{1}{2}gt^2 Using the vertical motion formula where y is the vertical position after time t, y_0 is the initial vertical position, g is the acceleration due to gravity.
2 y – y_0 = v_0 \sin(\theta) t – \frac{1}{2}gt^2 Rearranging the formula to isolate v_0 dependent terms on one side.
3 y_0 = 1.5 \, \text{m}, \, y = 12.2 \, \text{m} Substitute the initial height and the height as the ball passes over the wall since 9.0 m + 3.2 m = 12.2 m for the wall height and additional clearance.
4 12.2 = 1.5 + v_0 \sin(30^\circ) \left(\frac{45}{v_0 \cos(30^\circ)}\right) – \frac{1}{2} \times 9.8 \times \left(\frac{45}{v_0 \cos(30^\circ)}\right)^2 Substituting all values including t from part (a) into the equation.
5 v_0 can be solved from this equation numerically A numerical solution is required due to the complexity of the equation.

# Part (c): Find the velocity of the soccer ball as it passes over the wall.

Step Derivation/Formula Reasoning
1 v_{x_{final}} = v_0 \cos(\theta) Horizontal velocity remains unchanged as there is no air resistance.
2 v_{y_{final}} = v_0 \sin(\theta) – gt Final vertical velocity formula considering the effects of gravity over time.
3 v_{final} = \sqrt{v_{x_{final}}^2 + v_{y_{final}}^2} Magnitude of the final velocity using Pythagorean theorem.

# Part (d): If the same soccer ball is fired with the same initial speed at an angle of 55.5°, determine if it will clear the wall.

Step Derivation/Formula Reasoning
1 y = 1.5 + v_0 \sin(55.5^\circ) t – \frac{1}{2}gt^2 Using vertical motion formula to find height at wall for the new angle.
2 y \geq 9.0 Check if the computed y is greater than or equal to the height of the wall (without considering additional height).
3 Perform numerical calculations to determine y If y calculated is greater than 9 meters then the ball will clear the wall.

Each step can be computed to specific values using the given/derived v_0 and related time to reach the wall or over the wall for each scenario.

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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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