| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( x = v_x t \) | Determine the time \( t \) it takes for the ball to reach the catcher by using the formula for horizontal motion \( x \) where \( v_x \) is the horizontal velocity and \( x \) is the horizontal distance. |
| 2 | \( t = \frac{x}{v_x} \) | Rearrange the equation to solve for \( t \). The time \( t \) is the distance \( x \) divided by the horizontal velocity \( v_x \). |
| 3 | \( t = \frac{17.0 \, \text{m}}{41.0 \, \text{m/s}} \) | Substitute the given values into the equation. The horizontal distance \( x \) is \( 17.0 \, \text{m} \) and the horizontal velocity \( v_x \) is \( 41.0 \, \text{m/s} \). |
| 4 | \( t \approx 0.4146 \, \text{s} \) | Calculate the value of \( t \). The time it takes for the ball to reach the catcher is approximately \( 0.4146 \, \text{s} \). |
| 5 | \( \Delta y = \frac{1}{2} g t^2 \) | Determine the vertical drop \( \Delta y \) using the formula for vertical motion under gravity, where \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)). |
| 6 | \( \Delta y = \frac{1}{2} \cdot 9.8 \, \text{m/s}^2 \cdot (0.4146 \, \text{s})^2 \) | Substitute the known values into the equation. Use \( g = 9.8 \, \text{m/s}^2 \) and \( t \approx 0.4146 \, \text{s} \). |
| 7 | \( \Delta y \approx 0.84 \, \text{m} \) | Calculate the value of \( \Delta y \). The vertical drop is approximately \( 0.841 \, \text{m} \). |
| 8 | \( \Delta y \approx 0.84 \, \text{m} \) | The amount the baseball drops by the time it reaches the catcher is approximately \( 0.841 \, \text{m} \). |
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Which statements are not valid for a projectile? Take up as positive.
A ball is launched and lands \( 20 \) \( \text{m} \) away below the launch point \( 2.5 \) \( \text{s} \) later. The maximum height reached is \( 8.0 \) \( \text{m} \). What was the original launch velocity?
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.
In archery, should the arrow be aimed directly at the target? How should your angle of aim depend on the distance to the target? Explain without using equations.
A projectile is launched at \( 25 \) \( \text{m/s} \) at an angle of \( 37^{\circ} \). It lands on a platform that is \( 5.0 \) \( \text{m} \) above the launch height.
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 \( g \).
Which launch angle gives the greatest horizontal range, assuming level ground and no air resistance?
Measurements made in \(1910\) indicate that the common flea is an impressive jumper, given its size. Assume that a flea’s initial speed is \(2.1 \, \text{m/s}\), and that it leaps at an angle of \(21^\circ\) with respect to the horizontal. The jump lasts \(0.16 \, \text{s}\).
A soccer ball is kicked horizontally off an \( 85 \) \( \text{m} \) high cliff at a speed of \( 34 \) \( \text{m/s} \). What is the ball’s final speed when it hits the ground below?
A bald eagle in level flight at a height of \(135 \, \text{m}\) drops the fish it caught. If the eagle’s speed is \(25.0 \, \text{m/s}\) how far from the drop point will the fish land?
0.84 m
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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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