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| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[v_{x}^{2}=v_{i}^{2}+2(-g)\Delta x\] | Use the kinematic relation \(v_{x}^{2}=v_{i}^{2}+2a\Delta x\) with upward positive (so \(a=-g\)). \(v_{x}=14\,\text{m/s}\) at the window, \(\Delta x=18\,\text{m}\). |
| 2 | \[v_{i}^{2}=v_{x}^{2}+2g\Delta x\] | Algebraically solve for \(v_{i}^{2}\). |
| 3 | \[v_{i}^{2}=14^{2}+2(9.8)(18)=196+352.8=548.8\] | Substitute the numerical values. |
| 4 | \[\boxed{v_{i}=23.43\,\text{m/s}}\] | Take the square root to obtain the initial speed. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[0=v_{i}^{2}+2(-g)\Delta x_{\text{max}}\] | At the peak, the final velocity is zero, so set \(v_{x}=0\). |
| 2 | \[\Delta x_{\text{max}}=\frac{v_{i}^{2}}{2g}\] | Re-arrange for the upward displacement from the street. |
| 3 | \[\Delta x_{\text{max}}=\frac{548.8}{19.6}=28\,\text{m}\] | Insert \(v_{i}^{2}=548.8\) and \(g=9.8\,\text{m/s}^{2}\). |
| 4 | \[\boxed{28\,\text{m}}\] | The ball rises 28 m above the street. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[v_{x}=v_{i}-g t_{1}\] | Use \(v_{x}=v_{i}+at\) with \(a=-g\) to relate velocities and time. |
| 2 | \[t_{1}=\frac{v_{i}-v_{x}}{g}\] | Solve for \(t_{1}\), the interval from the throw to passing the window upward. |
| 3 | \[t_{1}=\frac{23.5-14}{9.8}=0.964\,\text{s}\] | Insert the numerical values. |
| 4 | \[\boxed{0.96\,\text{s}}\] | The ball was thrown roughly one second before being seen at the window. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[T_{\text{total}}=\frac{2v_{i}}{g}\] | For motion that starts and ends at the same height, total flight time is twice the time to the peak, \(v_{i}/g\). |
| 2 | \[T_{\text{total}}=\frac{2(23.5)}{9.8}=4.79\,\text{s}\] | Insert \(v_{i}=23.5\,\text{m/s}\) and \(g=9.8\,\text{m/s}^{2}\). |
| 3 | \[t_{\text{after window}}=T_{\text{total}}-t_{1}=4.79-0.96=3.83\,\text{s}\] | Subtract the elapsed time before passing the window to find the interval after it. |
| 4 | \[\boxed{4.8\,\text{s}\;\text{(total from throw)}}\] | The ball returns to the street 4.8 s after being thrown, i.e., about 3.8 s after passing the window. |
Just ask: "Help me solve this problem."
A tennis ball is thrown straight up with an initial speed of \( 22.5 \, \text{m/s} \). It is caught at the same distance above ground.
A 135.0 N force is applied to a 30.0 kg box at 42 degree angle to the horizontal. If the force of friction is 85.0, what is the net force and acceleration? If the object starts from rest, how far has it traveled in 3.3 sec?
A baseball is tossed from street level by a student straight up at a speed of \(25.3 \text{ m/s}\). After reaching maximum height, it is caught by another student on the roof of a building, \(17.4 \text{ m}\) above the street. How long did this take?
A car travels east at a steady \( 30 \) \( \text{m/s} \) for \( 5 \) \( \text{s} \). What is its acceleration during this motion?
An object is dropped from the top of a 45 m tall building
\(23.43\,\text{m/s}\)
\(28\,\text{m}\)
\(0.96\,\text{s}\)
\(4.8\,\text{s}\)
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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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