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| 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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Wile E. Coyote is (still) chasing after his arch-nemesis, the Roadrunner across a cliff that is \(125 \, \text{m}\) high. The Coyote is running in the horizontal direction towards the edge of a cliff when, at the last second, the Roadrunner steps out of the way and the witless coyote falls to the canyon floor.
An eagle is flying horizontally at \(6 \, \text{m/s}\) with a fish in its claws. It accidentally drops the fish.
Person A throws a ball horizontally from a cliff \( 20 \) \( \text{m} \) tall at \( 12 \) \( \text{m/s} \). Person B is running to the right on the ground and catches the ball at the same height it would’ve landed after running \( 15 \) \( \text{m} \). How fast was Person B running?
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 baseball rolls off a 0.70 m high desk and strikes the floor 0.25 m away from the base of the desk. How fast was the ball rolling?
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?
On a distant planet, golf is just as popular as it is on Earth. A golfer tees off and drives the ball \(3.5\) times as far as he would have on Earth, given the same initial velocities on both planets. The ball is launched at a speed of \(45 \, \text{m/s}\) at an angle of \(29^\circ\) above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet find:
A javelin thrower, of height \( 1.8 \) \( \text{m} \), throws a javelin with initial velocity of \( 26 \) \( \text{m s}^{-1} \) at \( 38^{\circ} \) to the horizontal. Calculate the time taken for the javelin to reach the ground from its maximum height. Give your answer in seconds and to an appropriate number of significant figures.
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 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?
\(1.77\,\text{s}\)
\(29.4\,\text{m/s}\)
\(25.6\,\text{m/s at }5.9^{\circ}\text{ below horizontal}\)
\(31.3\,\text{m}>9.0\,\text{m (clears)}\)
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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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