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UBQ Credits
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[v_{iy}=v_i\sin\theta\] | Resolve the initial speed into its vertical component \(v_{iy}\). |
| 2 | \[0 = v_{iy}^2 + 2(-g)\Delta y\] | At the peak the vertical velocity is zero; apply the kinematic equation with acceleration \(-g\). |
| 3 | \[\Delta y = \frac{v_{iy}^2}{2g}\] | Solve algebraically for the vertical displacement \(\Delta y\), the maximum height. |
| 4 | \[v_{iy}=36.6\sin42.2^\circ = 24.6\,\text{m/s}\] | Insert the given numbers to get \(v_{iy}\). |
| 5 | \[h_{\text{max}} = \frac{(24.6\,\text{m/s})^2}{2(9.80\,\text{m/s}^2)} = 30.9\,\text{m}\] | Calculate the numerical value of the height. |
| 6 | \[\boxed{30.9\,\text{m}}\] | Maximum height reached. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[t = \frac{2v_{iy}}{g}\] | Round-trip time is twice the time to reach the peak, using symmetry of the motion. |
| 2 | \[t = \frac{2(24.6\,\text{m/s})}{9.80\,\text{m/s}^2} = 5.02\,\text{s}\] | Substitute \(v_{iy}\) and \(g\). |
| 3 | \[\boxed{5.02\,\text{s}}\] | Total time in the air. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[v_x = v_i\cos\theta\] | Resolve the initial speed into its horizontal component \(v_x\). |
| 2 | \[v_x = 36.6\cos42.2^\circ = 27.1\,\text{m/s}\] | Insert the given numbers. |
| 3 | \[R = v_x t\] | The horizontal distance equals horizontal speed times total time (no horizontal acceleration). |
| 4 | \[R = 27.1\,\text{m/s}\times5.02\,\text{s} = 1.36\times10^{2}\,\text{m}\] | Compute the range. |
| 5 | \[\boxed{1.36\times10^{2}\,\text{m}}\] | Total horizontal distance. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[v_x = 27.1\,\text{m/s}\] | Horizontal speed remains constant throughout the flight. |
| 2 | \[v_y = v_{iy} – g t\] | Use the kinematic relation for vertical velocity after time \(t\). |
| 3 | \[v_y = 24.6\,\text{m/s} – (9.80\,\text{m/s}^2)(1.50\,\text{s}) = 9.9\,\text{m/s}\] | Insert the numbers to find \(v_y\) at 1.50 s. |
| 4 | \[v = \sqrt{v_x^2 + v_y^2}\] | Speed is the magnitude of the velocity vector. |
| 5 | \[v = \sqrt{(27.1\,\text{m/s})^2 + (9.9\,\text{m/s})^2} = 28.8\,\text{m/s}\] | Compute the magnitude. |
| 6 | \[\boxed{28.8\,\text{m/s}}\] | Speed 1.50 s after launch. |
Just ask: "Help me solve this problem."
A bird, traveling at \(50 \, \text{m/s}\) wants to hit a man \(100 \, \text{m}\) below with a dropping. How far in distance before flying directly over the man should the bird release it?
Barry Bonds hits a \(125 \,\text{m}\) home run. Assuming that the ball left the bat at an angle of \(45^\circ\) from the horizontal, calculate how long the ball was in the air.
An eagle is flying horizontally at \(6 \, \text{m/s}\) with a fish in its claws. It accidentally drops the fish.
You kick a soccer ball with an initial velocity directed 53° above the horizontal. The ball lands on a roof 7.2 m high. The wall of the building is 25 m away, and it takes the ball 2.1 seconds to pass directly over the wall.
Projectiles 1 and 2 are launched from level ground at the same time and follow the trajectories shown in the figure. Which one of the projectiles, if either, returns to the ground first, and why?
\(30.9\,\text{m}\)
\(5.02\,\text{s}\)
\(1.36\times10^{2}\,\text{m}\)
\(28.8\,\text{m/s}\)
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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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