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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v_f = v_i + at[/katex] | Use the kinematic equation to find the time to reach the maximum height. Here, [katex]v_f[/katex] is the final velocity (0 m/s), [katex]v_i[/katex] is the initial velocity (25.3 m/s), [katex]a[/katex] is the acceleration due to gravity (-9.8 m/s²), and [katex]t[/katex] is the time. |
| 2 | [katex]0 = 25.3 \, \text{m/s} – (9.8 \, \text{m/s}^2) t[/katex] | Substitute the known values into the kinematic equation. |
| 3 | [katex]t = \frac{25.3 \, \text{m/s}}{9.8 \, \text{m/s}^2}[/katex] | Solve for [katex]t[/katex], the time to reach maximum height. |
| 4 | [katex]t \approx 2.58 \, \text{s}[/katex] | Calculate the time numerically. |
| 5 | [katex]y_f = v_i t + \frac{1}{2}at^2[/katex] | Use the kinematic equation to find the total displacement from the initial point to the roof after maximum height. Here, [katex]y_f[/katex] is the final height (17.4 m), [katex]a[/katex] is the acceleration due to gravity (-9.8 m/s²), and [katex]v_i[/katex] is 0 since it starts falling from rest at maximum height. |
| 6 | [katex]17.4 = 0 + \frac{1}{2}(-9.8)t^2[/katex] | Substitute the values into the kinematic equation. We need to solve for the time [katex]t[/katex] it takes to fall back down 17.4 m. |
| 7 | [katex]17.4 = -4.9t^2[/katex] | Simplify the equation by calculating the constant in the kinematic equation. |
| 8 | [katex]t^2 = \frac{17.4}{4.9}[/katex] | Rearrange the equation to isolate [katex]t^2[/katex] on one side of the equation. |
| 9 | [katex]t = \sqrt{\frac{17.4}{4.9}}[/katex] | Solve for [katex]t[/katex]. |
| 10 | [katex]t \approx 1.88 \, \text{s}[/katex] | Calculate the time numerically for the descent from the maximum height to the roof level. |
| 11 | [katex] \mathbf{t_{\text{total}} = t_{\text{up}} + t_{\text{down}} = 2.58 \text{s} + 1.88 \text{s}} [/katex] | Add the time to reach the maximum height and the time to descend to the roof level. |
| 12 | [katex]\boxed{4.46 \text{s}}[/katex] | Final answer: the total time taken for the baseball to reach the roof level. |
Just ask: "Help me solve this problem."
A coin is dropped from a hot air-balloon that is 250 m above the ground rising at 11 m/s upwards. For the coin, assume up is positive and find the following:
An object is thrown straight upward at 64 m/s.
Which of the following statements about the acceleration due to gravity is TRUE?
Can an object’s average velocity equal zero when object’s speed is greater than zero? Explain using the formula for average velocity vs speed.
A 2,000 kg car collides with a stationary 1,000 kg car. Afterwards, they slide 6 m before coming to a stop. The coefficient of friction between the tires and the road is 0.7. Find the initial velocity of the 2,000 kg car before the collision?
4.46 seconds
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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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