| Derivation/Formula | Reasoning |
|---|---|
| \[\Delta y = \frac{1}{2} g t^{2}\] | Vertical motion equation with initial vertical speed equal to zero. |
| \[t = \sqrt{\frac{2\Delta y}{g}}\] | Solve algebraically for the fall time. |
| \[t = \sqrt{\frac{2 \times 85}{9.8}} = 4.16\,\text{s}\] | Insert \(\Delta y = 85\,\text{m}\) and \(g = 9.8\,\text{m/s}^{2}\). |
| \[v_{y} = g t\] | Final vertical speed gained during free fall. |
| \[v_{y} = 9.8 \times 4.16 = 40.8\,\text{m/s}\] | Numerical vertical velocity just before impact. |
| \[v_x = \sqrt{v_i^{2} + v_{y}^{2}}\] | Pythagorean combination of horizontal \(v_i\) and vertical \(v_{y}\) components. |
| \[v_x = \sqrt{34^{2} + 40.8^{2}} = 53.1\,\text{m/s}\] | Evaluate to find the resultant speed. |
| \[\boxed{v_x \approx 53\,\text{m/s}}\] | Final answer rounded to two significant figures. |
| \[41\,\text{m/s}\] | Option (b) considers only the vertical speed and ignores the horizontal component. |
| \[1681\,\text{m/s}\] | Option (c) is the square of 41 rather than the correct square-root operation. |
| \[776\,\text{m/s}\] | Option (d) results from multiplying instead of squaring and adding the speed components. |
| \[\text{More information?}\] | Option (e) is wrong because height and initial speed fully determine the final speed (mass and angle are unnecessary). |
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A gun can fire a bullet to height \( h \) when fired straight up. If the same gun is pointed at an angle of \( 45^\circ \) from the vertical, what is the new maximum height of the projectile?
A ball of mass \( M \) is attached to a string of length \( L \). It moves in a vertical circle and at the bottom the ball just clears the ground. The tension at the bottom of the path is \( 3 \) times the weight of the ball. Give all answers in terms of \( M \), \( L \), and \( g \).
One ball is dropped vertically from a window. At the same instant, a second ball is thrown horizontally from the same window. Which ball has the greater speed at ground level?
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 rock is thrown at an angle of \( 42^\circ \) above the horizontal at a speed of \( 14 \, \text{m/s} \). Determine how long it takes the rock to hit the ground.
Which of the following statements about the acceleration due to gravity is TRUE?
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.
An airplane with a speed of \( 97.5 \, \text{m/s} \) is climbing upward at an angle of \( 50.0^\circ \) with respect to the horizontal. When the plane’s altitude is \( 732 \, \text{m} \), the pilot releases a package.
A rifle is used to shoot a target twice, using identical cartridges. The first time, the rifle is aimed parallel to the ground and directly at the center of the bull’s-eye. The bullet strikes the target at a distance of \( H_A \) below the center, however. The second time, the rifle is similarly aimed, but from twice the distance from the target. This time the bullet strikes the target at a distance of \( H_B \) below the center. Find the ratio \( H_B / H_A \).
A ball of mass \( 0.5 \, \text{kg} \), initially at rest, is kicked directly toward a fence from a point \( 32 \, \text{m} \) away, as shown above. The velocity of the ball as it leaves the kicker’s foot is \( 20 \, \text{m/s} \) at an angle of \( 37^\circ \) above the horizontal. The top of the fence is \( 2.5 \, \text{m} \) high. The ball hits nothing while in flight and air resistance is negligible.
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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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