| Derivation/Formula | Reasoning |
|---|---|
| \[v_i = 22\ \text{m/s},\ \theta = 60^{\circ},\ \Delta y = -5\ \text{m}\] | Identify the given launch speed \(v_i\), launch angle \(\theta\), and vertical displacement \(\Delta y\), which is negative because the ramp is lower than the launch point. |
| \[v_{iy} = v_i \sin\theta = 22 \times \sin 60^{\circ}\] | Resolve the launch speed into its vertical component \(v_{iy}\). |
| \[v_{iy} = 22 \times \frac{\sqrt{3}}{2} = 19.052\ \text{m/s}\] | Use \(\sin 60^{\circ} = \tfrac{\sqrt{3}}{2} \approx 0.866\) to find the numerical value. |
| \[\Delta y = v_{iy} t + \frac{1}{2} a t^2\] | Apply the vertical kinematic equation, taking upward as positive and acceleration \(a = -g = -9.8\ \text{m/s}^2\). |
| \[-5 = 19.052 t – 4.9 t^2\] | Substitute \(\Delta y = -5\) and \(a = -9.8\ \text{m/s}^2\) into the kinematic equation. |
| \[4.9 t^2 – 19.052 t – 5 = 0\] | Rearrange the expression into standard quadratic form \(at^2 + bt + c = 0\). |
| \[t = \frac{19.052 \pm \sqrt{19.052^2 + 4(4.9)(5)}}{2 \times 4.9}\] | Use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\) with \(a = 4.9\), \(b = -19.052\), and \(c = -5\). |
| \[t = \frac{19.052 \pm 21.47}{9.8}\] | Compute the discriminant: \(19.052^2 + 4(4.9)(5) \approx 460.98\) and its square root \(\sqrt{460.98} \approx 21.47\). |
| \[t = 4.13\ \text{s}\] (\text{positive root}) | Choose the positive solution; the negative root corresponds to a non-physical time before launch. |
| \[\boxed{t = 4.13\ \text{s}}\] | Box the final numerical answer for the time of flight. |
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A rock is thrown from the top of a \( 15 \) \( \text{m} \) building at an unknown angle and speed. It hits a target on the ground \( 35 \) \( \text{m} \) away horizontally \( 3 \) \( \text{s} \) after launch. What was the rock’s launch angle?
A batter hits a fly ball which leaves the bat \( 0.90 \) \( \text{m} \) above the ground at an angle of \( 61^\circ \) with an initial speed of \( 28 \) \( \text{m/s} \) heading toward centerfield. Ignore air resistance.
A rock is dropped from the top of a tall tower. Half a second later another rock, twice as massive as the first, is dropped. Ignoring air resistance what is the correct choice?
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?
Two balls are thrown off a building with the same speed, one straight up and one at a 45° angle. Which statement is true if air resistance can be ignored?
Water balloons are tossed from the roof of a building, all with the same speed but with different launch angles. Which one has the highest speed when it hits the ground? Ignore air resistance. Without using equations, explain your answer.
A ball is shot from the top of a building with an initial velocity of \( 18 \) \( \text{m/s} \) at an angle \( \theta = 42^\circ \) above the horizontal.
A blue ball is thrown upward with a velocity of \( 9 \) \( \text{m/s} \) upward from the top of a high cliff. At the same time, a red ball is dropped from the same spot. The red ball is observed to hit the ground below exactly \( 1 \) \( \text{s} \) before the blue ball. How high is the cliff?
Two balls are launched at the same speed. Ball A is launched at an angle of \( 45^{\circ} \) and Ball B is launched at an angle of \( 60^{\circ} \). Which one reaches a higher point?
A projectile is fired with an initial speed of \( 36.6 \) \( \text{m/s} \) at an angle of \( 42.2^\circ \) above the horizontal on a long flat firing range.
\(t = 4.13\,\text{s}\)
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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] |
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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