# Solution for part (a) – Find the magnitude of the flea’s horizontal displacement.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( v_{x0} = v_0 \cos(\theta) \) | Determine the initial horizontal velocity component using the horizontal cosine component of the initial speed and launch angle. \( v_0 \) is the initial speed and \( \theta \) is the launch angle. |
| 2 | \( v_{x0} = 2.1 \cos(21^\circ) \) | Substitute \( v_0 = 2.1 \, \text{m/s} \) and \( \theta = 21^\circ \) into the horizontal velocity formula. |
| 3 | \( v_{x0} \approx 1.969 \, \text{m/s} \) | Calculate the initial horizontal velocity component. |
| 4 | \( x = v_{x0} \cdot t \) | Using the formula for horizontal displacement, where \( t \) is the total time of the jump. |
| 5 | \( x = 1.969 \cdot 0.16 \) | Insert \( v_{x0} \) and \( t = 0.16 \, \text{s} \) into the displacement formula. |
| 6 | \( x \approx 0.315 \, \text{m} \) | Compute the flea’s horizontal displacement. |
# Solution for part (b) – Find how high does the flea jump?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( v_{y0} = v_0 \sin(\theta) \) | Calculate the initial vertical velocity component using the vertical sine component of the initial speed and launch angle. |
| 2 | \( v_{y0} = 2.1 \sin(21^\circ) \) | Plug in \( v_0 = 2.1 \, \text{m/s} \) and \( \theta = 21^\circ \) into the vertical velocity formula. |
| 3 | \( v_{y0} \approx 0.752 \, \text{m/s} \) | Calculate the initial vertical velocity component. |
| 4 | \( y = \frac{v_{y0}^2}{2g} \) | The formula for vertical displacement due to gravity. Note the final vertical speed of the flea at maximum height is \( 0 \, m/s \). |
| 5 | \( y = \frac{0.752^2}{2 \times 9.81} \) | Substituting values. |
| 6 | \( y \approx 0.029 \, \text{m} \) | Calculate the peak height of the flea’s jump. |
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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.
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?
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 person shoots a basketball with a speed of \( 12 \, \text{m/s} \) at an angle of \( 35^\circ \) above the horizontal. If the person is \( 2.4 \, \text{m} \) tall and the hoop is \( 3.05 \, \text{m} \) above the ground, how far back must the person stand in order to make the shot?
A ball is tossed straight up while the thrower is standing in a moving train car that is moving at a constant velocity. Neglecting air resistance, what is the path of the ball relative to the ground outside the train?
A baseball is thrown at an angle of 25° relative to the ground at a speed of 23.0 m/s. The ball is caught 42.0 m from the thrower.
An eagle is flying horizontally at \(6 \, \text{m/s}\) with a fish in its claws. It accidentally drops the fish.
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.
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?
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?
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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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