| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_y = v\sin(\theta)\) | Resolve the initial velocity into its vertical component using the angle \( \theta = 45^{\circ} \). |
| 2 | \(v_y = v\sin(45^{\circ}) = v\frac{\sqrt{2}}{2}\) | Since \( \sin(45^{\circ}) = \frac{\sqrt{2}}{2} \). |
| 3 | \(h = \frac{v^2}{2g} \) | When fired straight up, the maximum height \( h \) is given by using the energy or kinematic equation. |
| 4 | \( h_{45} = \frac{(v \frac{\sqrt{2}}{2})^2}{2g} \) | Substitute the vertical component of the velocity at 45 degrees into the height formula. |
| 5 | \( h_{45} = \frac{v^2 \frac{2}{4}}{2g} = \frac{v^2}{4g} \) | Simplify the expression. |
| 6 | \(\frac{h}{2} = \frac{v^2}{2g} \rightarrow h_{45} = \frac{h}{2}\) | Since \(h = \frac{v^2}{2g} \), replace \( \frac{v^2}{2g}\) with \( h \). Therefore, \( h_{45} = \frac{h}{2} \). |
| 7 | (c) \( \frac{h}{2} \) | Thus, the final answer is option (c) \( \frac{h}{2} \). |
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A girl, standing still, tosses a ball vertically upwards. One second later, she tosses up another ball at the same velocity. The balls collide \( 0.5 \) \( \text{s} \) after the second ball is tossed. With what velocity were they tossed? The acceleration due to gravity is \( 9.8 \) \( \text{m/s}^2 \).
A kangaroo jumps straight up to a vertical height of \( 1.45 \) \( \text{m} \). How long was it in the air before returning to Earth?
A golfer hits a shot to a green that is elevated \(2.80 \, \text{m}\) above the point where the ball is struck. The ball leaves the club at a speed of \(18.9 \, \text{m/s}\) at an angle of \(52.0^\circ\) above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
At time \( t = 0 \) an object is traveling to the right along the \( +x \) axis at a speed of \( 10.0 \) \( \text{m/s} \) with acceleration \( -2.0 \) \( \text{m/s}^2 \). Which statement is true?
An object undergoes constant acceleration. Starting from rest, the object travels \( 5 \, \text{m} \) in the first second. Then it travels \( 15 \, \text{m} \) in the next second. What additional distance will be covered in the third second?
A golfer hits her ball in a high arcing shot. Air resistance is negligible. When the ball is at its highest point, which of the following is true?
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?
You throw a ball straight upward. It leaves your hand at \( 20 \) \( \text{m/s} \) and slows at a steady rate until it stops at the peak. The ball then comes back down, speeding up steadily until it hits the ground with the same speed it left your hand. Draw the velocity vs. time graph or explain it in terms of functions.
A ball is thrown horizontally from the roof of a building \( 7.5 \) \( \text{m} \) tall and lands \( 9.5 \) \( \text{m} \) from the base. What was the ball’s initial speed?
An object is thrown straight upward at 64 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] |
Metric Prefixes
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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