| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( \Delta x = v_i \cos( \theta) t \) | Use the horizontal motion equation. This relates displacement to initial velocity, angle, and time. Here, \(\theta = 45°\) and \(\Delta x = 125 \, \text{m}\). |
| 2 | \( y = v_i \sin( \theta) t – \frac{1}{2} g t^2 \) | Use the vertical motion equation. This relates height to initial velocity, angle, time, and gravitational acceleration \( g \approx 9.8 \, \text{m/s}^2 \). |
| 3 | \( y = 0 \, \text{m} \) | At landing, the height is assumed to be zero since the baseball is hit and lands at the same vertical level. |
| 4 | \( 0 = v_i \sin( \theta) t – \frac{1}{2} g t^2 \) | Substitute \( y = 0 \) into the vertical motion equation. |
| 5 | \( t = 0 \) or \( t = \frac{2 v_i \sin( \theta)}{ g} \) | Solving the quadratic equation, we discard the trivial solution \( t = 0 \) and use \( t = \frac{2 v_i \sin( \theta)}{ g} \). |
| 6 | \( \Delta x = v_i \cos( \theta) t \) | Substitute \( t \) into the horizontal motion equation and solve for \( v_i \). |
| 7 | \( 125 = v_i \cos( 45°) \frac{2 v_i \sin( 45°)}{ g} \) | Replace \( t \) with the expression found and simplify using \(\cos(45°) = \sin(45°) = \frac{\sqrt{2}}{2} \). |
| 8 | \( 125 = \frac{v_i^2 \cdot 2 (\frac{\sqrt{2}}{2})^2}{ 9.8} \) | Simplify the trigonometric expressions. |
| 9 | \( 125 = \frac{v_i^2}{ 9.8} \) | Simplify further and solve for \( v_i \). |
| 10 | \( v_i = \sqrt{125 \cdot 9.8 } \approx 35 \, \text{m/s} \) | Calculate the initial velocity \( v_i \). |
| 11 | \( t = \frac{2 \cdot 35 \cdot \frac{\sqrt{2}}{2}}{ 9.8} \approx 5 \, \text{s} \) | Substituting \(v_i \) into the time equation to find the total time of flight |
| 12 | \(\boxed{5 \, \text{s}}\) | The total time of flight for the ball was 5 seconds. |
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A cylindrical tank of water (height \( H \)) is punctured at a height \( h \) above the bottom. How far from the base of the tank will the water stream land (in terms of \( h \) and \( H \))? What must the value of \( h \) be such that the distance at which the stream lands will be equal to \( H \)?
Two balls are launched at the same time from opposite sides of a \( 100 \) \( \text{m} \) wide and \(1000 ~\text{m}\) canyon. Ball A is launched at \( 20 \) \( \text{m/s} \) at \( 45^{\circ} \) from the left side. Ball B is launched at \( 20 \) \( \text{m/s} \) at \( 45^{\circ} \) from the right side.
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?
If a baseball pitch leaves the pitcher’s hand horizontally at a velocity of \( 150 \) \( \text{km/h} \), by what \( \% \) will the pull of gravity change the magnitude of the velocity when the ball reaches the batter, \( 18 \) \( \text{m} \) away? For this estimate, ignore air resistance and spin on the ball.
A projectile is launched at \( 20 \) \( \text{m/s} \) and lands \( 35 \) \( \text{m} \) away on level ground. At what two horizontal positions is the projectile exactly \( 5.0 \) \( \text{m} \) above the ground?
A projectile is launched at an angle of \( 30^{\circ} \) and hits a vertical wall \( 40 \) \( \text{m} \) away. After bouncing back horizontally, it lands \( 15 \) \( \text{m} \) behind the launch point. How high up on the wall did the projectile strike?
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.
Two cannonballs, A and B, are fired from the ground with identical initial speeds, but with \( \theta_A \) larger than \( \theta_B \).
A block of mass \(M_1\) travels horizontally with a constant speed \(v_0\) on a plateau of height \(H\) until it comes to a cliff. A toboggan of mass \(M_2\) is positioned on level ground below the cliff. The center of the toboggan is a distance \(D\) from the base of the cliff.
A soccer ball with an initial height of \(1.5 \, \text{m}\) above the ground is launched at an angle of \(30^\circ\) above the horizontal. The soccer ball travels a horizontal distance of \(45 \, \text{m}\) to a \(9.0 \, \text{m}\) high castle wall, and passes over \(3.20 \, \text{m}\) above the highest point of the wall. Assume air resistance is negligible.
5 seconds
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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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