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| Derivation/Formula | Reasoning |
|---|---|
| \[v_i\sin\theta = 25(0.602)=15.0\,\text{m/s}\] | Resolve the launch speed into the vertical component using \(\sin37^\circ\approx0.602\). |
| \[\Delta y = v_i\sin\theta\,t – \tfrac12 g t^2\] | Use the vertical displacement equation with \(\Delta y = +5.0\,\text{m}\) and \(g=9.8\,\text{m/s}^2\). |
| \[4.9t^2-15.0t+5.0=0\] | Substitute numerical values and rearrange to standard quadratic form. |
| \[t=\frac{15.0\pm11.3}{9.8}\] | Apply the quadratic formula: \(t=\tfrac{-b\pm\sqrt{b^2-4ac}}{2a}\). |
| \[t=0.38\,\text{s}\quad\text{or}\quad t=2.69\,\text{s}\] | The smaller root is when the projectile is still rising; the larger root is the total flight time. |
| \[\boxed{t=2.69\,\text{s}}\] | Total time aloft. |
| Derivation/Formula | Reasoning |
|---|---|
| \[v_i\cos\theta = 25(0.799)=20.0\,\text{m/s}\] | Resolve the launch speed into the constant horizontal component using \(\cos37^\circ\approx0.799\). |
| \[\Delta x = v_i\cos\theta\;t\] | Horizontal displacement equals horizontal velocity times time; no horizontal acceleration. |
| \[\Delta x = 20.0\times2.69 = 5.38\times10^{1}\,\text{m}\] | Insert \(t=2.69\,\text{s}\) from part (a). |
| \[\boxed{\Delta x = 53.8\,\text{m}}\] | Horizontal range. |
| Derivation/Formula | Reasoning |
|---|---|
| \[v_{y\,(\text{top})}=0\,\text{m/s}\] | At the highest point the vertical component of velocity is momentarily zero for any projectile motion neglecting air resistance. |
| \[\boxed{0\,\text{m/s}}\] | Result. |
| Derivation/Formula | Reasoning |
|---|---|
| \[v_{y\,(\text{land})}=v_i\sin\theta – g t = 15.0 – 9.8(2.69) = -11.3\,\text{m/s}\] | Vertical velocity after \(t=2.69\,\text{s}\); negative sign indicates downward direction. |
| \[v_{x}=20.0\,\text{m/s}\] | Horizontal velocity is unchanged throughout flight. |
| \[v_x^2 = v_{y\,(\text{land})}^2 + (v_{x})^2 = (11.3)^2 + (20.0)^2 = 5.27\times10^{2}\] | Pythagorean addition of components for speed magnitude (labelled here as \(v_x\) per notation). |
| \[v_x = 22.9\,\text{m/s}\] | Square-root of previous result. |
| \[\boxed{v_x < v_i}\] | The landing speed \(22.9\,\text{m/s}\) is less than the launch speed \(25\,\text{m/s}\) because the projectile finishes 5 m higher, converting some kinetic energy into gravitational potential energy. |
Just ask: "Help me solve this problem."
A projectile has the least speed at what point in its path?
A projectile is launched at an upward angle of \( 30^\circ \) to the horizontal with a speed of \( 30 \) \( \text{m/s} \). How does the horizontal component of its velocity \( 1.0 \) \( \text{s} \) after launch compare with its horizontal component of velocity \( 2.0 \) \( \text{s} \) after launch, ignoring air resistance?
A rubber ball bounces on the ground. After each bounce, the ball reaches one-half the height of the bounce before it. If the time the ball was in the air between the first and second bounce was 1 second. What would be the time between the second and third bounce?
A plane, 220 meters high, is dropping a supply crate to an island below. It is traveling with a horizontal velocity of 150 m/s. At what horizontal distance must the plane drop the supply crate for it to land on the island? Use [katex] g = 9.81 \, m/s^2[/katex].
You are a bungee jumping fanatic and want to be the first bungee jumper on Jupiter. The length of your bungee cord is \( 45.0 \) \( \text{m} \). Free fall acceleration on Jupiter is \( 23.1 \) \( \text{m/s}^2 \). What is the ratio of your speed on Jupiter to your speed on Earth when you have dropped \( 45 \) \( \text{m} \)? Ignore the effects of air resistance and assume that you start at rest.
\(2.69\,\text{s}\)
\(53.8\,\text{m}\)
\(0\,\text{m/s}\)
\(v_x < v_i\)
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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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