# 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.
Phy can also check your working. Just snap a picture!
A javelin thrower standing at rest holds the center of the javelin behind her head, then accelerates it through a distance of 70 cm as she throws. She releases the 600 g javelin 2.0 m above the ground traveling at an angle of 30° above the horizontal. In this throw, the javelin hits the ground 54 m away. Find the following:
Measurements made in 1910 indicate that the common flea is an impressive jumper, given its size. Assume that a flea’s initial speed is 2.1 m/s, and that it leaps at an angle of 21° with respect to the horizontal. The jump lasts 0.16 s.
A arrow is shot horizontally from a distance of 20 meters away. It lands .05 meters below the center of the target. If air resistance is negligible what was the initial speed of the arrow?
In a lab experiment, a ball is rolled down a ramp so that it leaves the edge of the table with a horizontal velocity v. Assume there are no frictional forces. If the table has a height h above the ground, how far away from the edge of the table, a distance x, does the ball land?
A cat chases a mouse across a 1.0 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 2.2 m from the edge of the table. When the cat slid off the table, what was its speed?
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| Kinematics | Forces |
|---|---|
| \Delta x = v_i t + \frac{1}{2} at^2 | F = ma |
| v = v_i + at | F_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 Motion | Energy |
|---|---|
| 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 |
| Momentum | Torque and Rotations |
|---|---|
| p = m v | \tau = r \cdot F \cdot \sin(\theta) |
| J = \Delta p | I = \sum mr^2 |
| p_i = p_f | L = I \cdot \omega |
| Simple Harmonic Motion |
|---|
| F = -k x |
| T = 2\pi \sqrt{\frac{l}{g}} |
| T = 2\pi \sqrt{\frac{m}{k}} |
| Constant | Description |
|---|---|
| g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
| G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
| \mu_k and \mu_s | Coefficients 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. |
| k | Spring 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 |
| Variable | SI 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)} |
| Variable | Derived 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)} |
| \omega (Angular Velocity) | \text{radians per second (rad/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.
Start with the given measurement: \text{5 km}
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}}
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}
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} |
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