| Derivation/Formula | Reasoning |
|---|---|
| \[ \Delta x = v_i \cos 30^{\circ}\, t \] | Horizontal motion: distance \( \Delta x = 45\,\text{m} \) equals horizontal speed times time. |
| \[ t = \frac{\Delta x}{v_i \cos 30^{\circ}} \] | Solve previous relation algebraically for time \( t \). |
| \[ t = \frac{45}{29.4\cos 30^{\circ}} \approx 1.77\,\text{s} \] | Insert the value of \( v_i \) found in part (b) to evaluate the time. |
| \[ \boxed{t \approx 1.77\,\text{s}} \] | Time taken for the ball to reach the wall. |
| Derivation/Formula | Reasoning |
|---|---|
| \[ y = y_0 + v_i \sin 30^{\circ}\, t – \tfrac12 g t^{2} \] | Vertical position of the ball; here \( y_0 = 1.5\,\text{m} \) and final \( y = 12.2\,\text{m} \). |
| \[ 12.2 = 1.5 + v_i \sin 30^{\circ}\, t – 4.9 t^{2} \] | Substitute heights and \( g = 9.8\,\text{m/s}^2 \). |
| \[ 10.7 = 0.5 v_i t – 4.9 t^{2} \] | Simplify numerical terms. |
| \[ t = \frac{45}{v_i \cos 30^{\circ}} \] | Replace \( t \) using the horizontal relation from part (a). |
| \[ 10.7 = 0.5 v_i \left(\frac{45}{v_i \cos 30^{\circ}}\right) – 4.9 \left(\frac{45}{v_i \cos 30^{\circ}}\right)^2 \] | Substitute the expression for \( t \) into the vertical equation. |
| \[ v_i \approx 29.4\,\text{m/s} \] | Algebraic manipulation (no calculus) gives the speed; evaluate numerically. |
| \[ \boxed{v_i \approx 29.4\,\text{m/s}} \] | Initial launch speed of the soccer ball. |
| Derivation/Formula | Reasoning |
|---|---|
| \[ v_x = v_i \cos 30^{\circ} \] | Horizontal component remains constant. |
| \[ v_x \approx 25.5\,\text{m/s} \] | Insert \( v_i = 29.4\,\text{m/s} \). |
| \[ v_y = v_i \sin 30^{\circ} – g t \] | Vertical component at the wall. |
| \[ v_y \approx -2.62\,\text{m/s} \] | Negative sign indicates downward motion. |
| \[ v = \sqrt{v_x^{2} + v_y^{2}} \] | Magnitude of the resultant velocity vector. |
| \[ v \approx 25.6\,\text{m/s} \] | Numerical magnitude. |
| \[ \phi = \tan^{-1}\!\left(\frac{|v_y|}{v_x}\right) \approx 5.9^{\circ} \] | Angle the velocity makes with the horizontal; below because \( v_y < 0 \). |
| \[ \boxed{v \approx 25.6\,\text{m/s\, at }5.9^{\circ}\text{ below horizontal}} \] | Complete velocity description as it clears the wall. |
| Derivation/Formula | Reasoning |
|---|---|
| \[ t_2 = \frac{\Delta x}{v_i \cos 55.5^{\circ}} \] | Time to reach the wall with the higher angle. |
| \[ t_2 \approx 2.70\,\text{s} \] | Evaluate using \( v_i = 29.4\,\text{m/s} \). |
| \[ y_2 = y_0 + v_i \sin 55.5^{\circ} t_2 – \tfrac12 g t_2^{2} \] | Vertical position at the wall for the new angle. |
| \[ y_2 \approx 31.3\,\text{m} \] | Numerical height above ground. |
| \[ \boxed{y_2 > 9.0\,\text{m} \;\text{(clears wall)}} \] | The ball still passes far above the castle wall. |
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A golfer hits her ball in a high arcing shot. Air resistance is negligible. When the ball is at its highest point, which of the following is true?
Suppose the water at the top of Niagara Falls has a horizontal speed of \( 2.7 \, \text{m/s} \) just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a \( 75^\circ \) angle below the horizontal?
A diver springs upward from a diving board. At the instant she contacts the water, her speed is \( 8.90 \, \text{m/s} \), and her body is extended at an angle of \( 75.0^\circ \) with respect to the horizontal surface of the water. At this instant, her vertical displacement is \( -3.00 \, \text{m} \), where downward is the negative direction. Determine her initial velocity, both magnitude and direction.
A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.0° above the horizontal. The rocket is fired toward an 11.0-m-high wall, which is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?
Three identical rocks are launched with identical speeds from the top of a platform of height \( h_0 \).
Which of the following correctly relates the magnitude \( v_y \) of the vertical component of the velocity of each rock immediately before it hits the ground?
Seo-Jun throws a ball to her friend Zuri. The ball leaves Seo-Jun’s hand from a height \( h = 1.5 \) \( \text{m} \) above the ground with an initial speed \( \vec{v}_{s,0} = 12 \) \( \text{m/s} \) at an angle of \( \theta = 25^\circ \) with respect to the horizontal. Zuri catches the ball at a height of \( h = 1.5 \) \( \text{m} \) above the ground.
After catching the ball, Zuri throws it back to Seo-Jun. The ball leaves Zuri’s hand from a height \( h = 1.5 \) \( \text{m} \) above the ground. The ball is moving with a speed of \( 15 \) \( \text{m/s} \) when it reaches a maximum height of \( 5.8 \) \( \text{m} \) above the ground.
At what height \( h’ \) above the ground will the ball be when the return throw reaches Seo-Jun?
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 \( g = 9.81 \, m/s^2\).
Barry Bonds hits a \(125 \,\text{m}\) home run. Assuming that the ball left the bat at an angle of \(45^\circ\) from the horizontal, calculate how long the ball was in the air.
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 textbook is launched up with a speed of 20 m/s, at an angle of 36°, from a 12 m high roof.
\(1.77\,\text{s}\)
\(29.4\,\text{m/s}\)
\(25.6\,\text{m/s at }5.9^{\circ}\text{ below horizontal}\)
\(31.3\,\text{m}>9.0\,\text{m (clears)}\)
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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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