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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]y = \frac{1}{2} g t^2[/katex] | Since the ball is in free fall, its vertical motion is described by the kinematic formula for position under constant acceleration, where [katex]y[/katex] is the vertical displacement, [katex]g[/katex] is the acceleration due to gravity (approximately [katex]9.81 \, \text{m/s}^2[/katex]) and [katex]t[/katex] is the time of fall. |
| 2 | [katex]t = \sqrt{\frac{2y}{g}}[/katex] | To find the time it takes for the ball to reach the ground, solve for [katex]t[/katex] from the vertical displacement equation, using [katex]y = 20 \, \text{m}[/katex] for the height of the cliff. |
| 3 | [katex]v_y = g t[/katex] | Use the vertical velocity formula where [katex]v_y[/katex] is the velocity in the vertical direction at time [katex]t[/katex]. |
| 4 | [katex]v_y = g \sqrt{\frac{2y}{g}}[/katex] | Substitute the expression for [katex]t[/katex] from step 2 into the formula for [katex]v_y[/katex] to express the vertical velocity as a function of [katex]y[/katex] and [katex]g[/katex]. |
| 5 | [katex]v_y = \sqrt{2gy}[/katex] | Simplify the expression for [katex]v_y[/katex]. |
| 6 | [katex]v_y = \sqrt{2 \cdot 9.81 \, \text{m/s}^2 \cdot 20 \, \text{m}}[/katex] | Plug in the values for [katex]g[/katex] and [katex]y[/katex] to calculate [katex]v_y[/katex]. |
| 7 | [katex]v_y = 19.8 \, \text{m/s}[/katex] | Calculate the numerical value of [katex]v_y[/katex]. |
| 8 | [katex]v = \sqrt{v_x^2 + v_y^2}[/katex] | The total velocity [katex]v[/katex] right before hitting the ground is found using the Pythagorean theorem, where [katex]v_x[/katex] is the horizontal velocity. |
| 9 | [katex]v = \sqrt{11^2 + 19.8^2}[/katex] | Given the horizontal velocity [katex]v_x = 11 \, \text{m/s}[/katex], use the vertical velocity calculated previously. |
| 10 | [katex]v = 22.7 \, \text{m/s}[/katex] | Calculate the final total velocity of the ball just before it hits the ground. |
Just ask: "Help me solve this problem."
A rock is thrown at an angle of \( 42^\circ \) above the horizontal at a speed of \( 14 \, \text{m/s} \). Determine how long it takes the rock to hit the ground.
You must split an apple resting on top of you friend’s head from a distance of 27 m. When you aim directly at the apple, the arrow is horizontal. At what angle should you aim the arrow to hit the apple if the arrow travels at a speed of 35 m/s?
In a lab experiment, a ball is rolled down a ramp so that it leaves the edge of the table with a horizontal velocity [katex]v[/katex]. Assume there are no frictional forces. If the table has a height [katex]h[/katex] above the ground, how far away from the edge of the table, a distance [katex]x[/katex], does the ball land?
A marble is thrown horizontally with a speed of 15 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of 65° with the horizontal. From what height above the ground was the marble thrown?
An arrow is shot horizontally from a distance of \( 20 \, \text{m} \) away. It lands \( 0.05 \, \text{m} \) below the center of the target. If air resistance is negligible, what was the initial speed of the arrow?
22.7 m/s
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [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 |
| Variable | SI 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] |
| Variable | Derived 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]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/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.
Start with the given measurement: [katex]\text{5 km}[/katex]
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]
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]
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] |
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