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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."
During projectile motion (neglecting air resistance), what is the vertical acceleration at the highest point, assuming the initial velocity is upwards in the positive direction?
Which of the following statements about the acceleration due to gravity is TRUE?
A baseball rolls off a 0.70 m high desk and strikes the floor 0.25 m away from the base of the desk. How fast was the ball rolling?
A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge that is \( 235 \) \( \text{m} \) below. The plane is traveling horizontally with a speed of \( 250 \) \( \text{km/h} \). How far in advance of the recipients (horizontal distance) must the goods be dropped?
Two cannonballs, A and B, are fired from the ground with identical initial speeds, but with \( \theta_A \) larger than \( \theta_B \).
\(2.69\,\text{s}\)
\(53.8\,\text{m}\)
\(0\,\text{m/s}\)
\(v_x < v_i\)
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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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