| 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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In a lab experiment, a ball is rolled down a ramp so that it leaves the edge of the table with a horizontal velocity \(v\). Assume there are no frictional forces. If the table has a height \(h\) above the ground, how far away from the edge of the table, a distance \(x\), does the ball land?
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.
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 ball is thrown horizontally from the roof of a building \( 7.5 \) \( \text{m} \) tall and lands \( 9.5 \) \( \text{m} \) from the base. What was the ball’s initial speed?
A projectile is launched at an upward angle of \( 30^\circ \) to the horizontal with a speed of \( 30 \) \( \text{m/s} \). How does the horizontal component of its velocity \( 1.0 \) \( \text{s} \) after launch compare with its horizontal component of velocity \( 2.0 \) \( \text{s} \) after launch, ignoring air resistance?
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.
A stone is thrown horizontally at \(8.0 \, \text{m/s}\) from a cliff \(80 \,\text{m}\) high. How far from the base of the cliff will the stone strike the ground?
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 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?
Which of the following statements about the acceleration due to gravity is TRUE?
\(1.77\,\text{s}\)
\(29.4\,\text{m/s}\)
\(25.6\,\text{m/s at }5.9^{\circ}\text{ below horizontal}\)
\(31.3\,\text{m}>9.0\,\text{m (clears)}\)
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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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