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# Part (a): At what times is the object 20 m above the ground?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]y = v_0 t + \frac{1}{2} a t^2[/katex] | Use the kinematic equation for vertical displacement where [katex]y[/katex] is the height, [katex]v_0[/katex] is the initial velocity, [katex]t[/katex] is the time, and [katex]a[/katex] is the acceleration. |
| 2 | [katex]20 = 64 t – \frac{1}{2} \cdot 9.8 \cdot t^2[/katex] | Substitute [katex]y = 20 \text{ m}[/katex], [katex]v_0 = 64 \text{ m/s}[/katex], and [katex]a = -9.8 \text{ m/s}^2[/katex]. Note the negative sign of acceleration due to gravity. |
| 3 | [katex]4.9 t^2 – 64 t + 20 = 0[/katex] | Rearrange the equation into standard quadratic form [katex]a t^2 + b t + c = 0[/katex] where [katex]a = 4.9[/katex], [katex]b = -64[/katex], and [katex]c = 20[/katex]. |
| 4 | [katex]t = \frac{64 \pm \sqrt{(-64)^2 – 4 \cdot 4.9 \cdot 20}}{2 \cdot 4.9}[/katex] | Use the quadratic formula [katex]t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}[/katex] to solve for [katex]t[/katex]. |
| 5 | [katex]t_1 = \frac{64 + 60.88}{9.8} \approx 12.74 \text{ s}, \\ t_2 = \frac{64 – 60.88}{9.8} \approx 0.32 \text{ s}[/katex] | Calculate the two possible values for [katex]t[/katex]. The results are [katex]t_1 \approx 12.74 \text{ s}[/katex] and [katex]t_2 \approx 0.32 \text{ s}[/katex]. |
| 6 | [katex]t = 0.32 \text{ s}, 12.74 \text{ s}[/katex] | Final answer: There are two times when the object is 20 m above the ground, at [katex]0.32 \text{ s}[/katex] and [katex]12.74 \text{ s}[/katex]. |
# Part (b): Why are there two answers for part (a)?
| Step | Explanation | Reasoning |
|---|---|---|
| 1 | Two Answers Explanation | There are two answers because the object passes 20 m twice: once while going up and once while coming back down. |
# Part (c): How long does it take to come back to the ground?
| Step | Derivation/Formula/Answer | Reasoning |
|---|---|---|
| 1 | [katex]y = v_0 t + \frac{1}{2} a t^2[/katex] | Use the kinematic equation for vertical displacement where [katex]y[/katex] is zero as it returns to the ground. |
| 2 | [katex]0 = 64 t – \frac{1}{2} \cdot 9.8 \cdot t^2[/katex] | Substitute [katex]y = 0 \text{ m}[/katex], [katex]v_0 = 64 \text{ m/s}[/katex], and [katex]a = -9.8 \text{ m/s}^2[/katex]. |
| 3 | [katex]0 = 64 t – 4.9 t^2[/katex] | Simplify the equation by calculating [katex]\frac{1}{2} \cdot 9.8 = 4.9[/katex]. |
| 4 | [katex]0 = t (64 – 4.9t)[/katex] | Factor out [katex]t[/katex] to solve for the time at which the object returns to the ground. |
| 5 | [katex]t = 0 \quad \text{or} \quad 64 = 4.9t \Rightarrow t = \frac{64}{4.9} \approx 13.06 \text{ s}[/katex] | The first solution [katex]t = 0[/katex] corresponds to the initial throw. Solving [katex]64 – 4.9t = 0[/katex] gives the time it takes to reach the ground. |
| 6 | [katex]t = 13.06 \text{ s}[/katex] | Final answer: The object takes [katex]13.06 \text{ s}[/katex] to come back to the ground. |
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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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