(a) How far does the textbook travel horizontally after it is released?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v_{0x} = v_0 \cos(\theta)[/katex] | Calculate the initial horizontal velocity. Use the initial speed and the angle of projection. |
| 2 | [katex]v_{0y} = v_0 \sin(\theta)[/katex] | Calculate the initial vertical velocity. Use the initial speed and the angle of projection. |
| 3 | [katex]v_{0x} = 20 \cos(36^\circ) \approx 16.2 \, \text{m/s}[/katex] | Substitute [katex] v_0 = 20 \, \text{m/s} [/katex] and [katex] \theta = 36^\circ [/katex] into the horizontal velocity formula. |
| 4 | [katex]v_{0y} = 20 \sin(36^\circ) \approx 11.8 \, \text{m/s}[/katex] | Substitute [katex] v_0 = 20 \, \text{m/s} [/katex] and [katex] \theta = 36^\circ [/katex] into the vertical velocity formula. |
| 5 | [katex]y = v_{0y} t – \frac{1}{2} g t^2 + \text{initial height}[/katex] | Use the equation of motion in the vertical direction. The textbook is moving under gravity. |
| 6 | [katex]0 = 12 + 11.8 t – \frac{1}{2} \cdot 9.8 t^2[/katex] | Set the displacement [katex] y [/katex] to zero because we are calculating the time [katex] t [/katex] when the textbook reaches the ground. [katex] g = 9.8 \text{m/s}^2 [/katex]. |
| 7 | [katex]4.9 t^2 – 11.8 t – 12 = 0[/katex] | Simplify the quadratic equation to solve for [katex] t [/katex]. |
| 8 | [katex] t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} [/katex] | Use the quadratic formula where [katex] a = 4.9 [/katex], [katex] b = -11.8 [/katex], and [katex] c = -12 [/katex]. |
| 9 | [katex] t \approx 3.18 \, \text{s} [/katex] | Solve the equation and take the positive root. This is the time the textbook stays in the air. |
| 10 | [katex] x = v_{0x} \cdot t [/katex] | Calculate the horizontal distance the textbook travels. Use the horizontal velocity and the time. |
| 11 | [katex] x \approx 16.2 \times 3.18 \approx 51.5 \, \text{m} [/katex] | Substitute [katex] v_{0x} = 16.2 \, \text{m/s} [/katex] and [katex] t = 3.18 \, \text{s} [/katex] into the horizontal distance formula to get the final answer. |
| [katex] \text{The horizontal distance traveled is approximately } 51.5 \, \text{m}[/katex] | ||
(b) What is the book’s velocity (speed and direction) when it reaches the ground?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v_y = v_{0y} – g t [/katex] | Calculate the final vertical velocity using the initial vertical velocity, gravitational acceleration, and time. |
| 2 | [katex]v_y = 11.8 – 9.8 \times 3.18 \approx -19.4 \, \text{m/s}[/katex] | Substitute [katex] v_{0y} = 11.8 \, \text{m/s} [/katex], [katex] g = 9.8 \, \text{m/s}^2 [/katex], and [katex] t = 3.18 \, \text{s} [/katex] into the vertical velocity formula. |
| 3 | [katex]v_{\text{total}} = \sqrt{v_{x}^2 + v_y^2} [/katex] | Calculate the magnitude of the total velocity using the Pythagorean theorem. |
| 4 | [katex]v_{\text{total}} \approx \sqrt{16.2^2 + (-19.4)^2} \approx 25.3 \, \text{m/s} [/katex] | Substitute [katex] v_{x} = 16.2 \, \text{m/s} [/katex] and [katex] v_y = -19.4 \, \text{m/s} [/katex] into the total velocity formula. |
| 5 | [katex]\theta = \tan^{-1} \left(\frac{v_y}{v_x}\right) [/katex] | Calculate the direction of the velocity. Use the inverse tangent to find the angle. |
| 6 | [katex]\theta \approx \tan^{-1} \left(\frac{-19.4}{16.2}\right) \approx -50.1^\circ [/katex] | The vector points 50.1° below the x-axis. |
| [katex] \text{The velocity when the book reaches the ground is approximately } 25.3 \, \text{m/s} \text{ at } -50.1^\circ [/katex] | ||
(c) What is the book’s maximum height above the ground?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v_y = 0 [/katex] | The vertical velocity at the maximum height is zero. |
| 2 | [katex]v_y = v_{0y} – g t[/katex] | Use the vertical motion equation to find the time to reach maximum height. |
| 3 | [katex]0 = 11.8 – 9.8 t[/katex] | Set final vertical velocity [katex] v_y = 0 [/katex] and solve for [katex] t [/katex]. |
| 4 | [katex]t = \frac{11.8}{9.8} \approx 1.20 \, \text{s}[/katex] | Solving the equation gives the time to reach maximum height. |
| 5 | [katex]H = v_{0y} t – \frac{1}{2} g t^2 + \text{initial height}[/katex] | Use the vertical motion equation to find the maximum height. |
| 6 | [katex]H \approx 11.8 \times 1.2 – \frac{1}{2} \times 9.8 \times (1.2)^2 + 12 [/katex] | Substitute [katex] v_{0y} = 11.8 \, \text{m/s} [/katex], [katex] g = 9.8 \, \text{m/s}^2 [/katex], [katex] t = 1.2 \, \text{s} [/katex], and initial height = 12 m. |
| 7 | [katex]H \approx 19.1 \, \text{m}[/katex] | Calculate the maximum height above the ground. |
| [katex] \text{The maximum height above the ground is approximately } 19.1 \, \text{m} [/katex] | ||
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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?
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 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 projectile is launched at a speed of \( 22 \) \( \text{m/s} \) at an angle of \( 60^{\circ} \) above the horizontal. It lands on a ramp that is \( 5 \) \( \text{m} \) lower than the launch height. How long does it take for the projectile to hit the ramp?
A rocket-powered hockey puck has a thrust of \(4.40 \, \text{N}\) and a total mass of \(1.00 \, \text{kg}\). It is released from rest on a frictionless table, \(2.10 \, \text{m}\) from the edge of a \(2.10 \, \text{m}\) drop. The front of the rocket is pointed directly toward the edge. Assuming that the thrust of the rocket is present for the entire time of travel, how far does the puck land from the base of the table?
A ball of mass \( 0.5 \, \text{kg} \), initially at rest, is kicked directly toward a fence from a point \( 32 \, \text{m} \) away, as shown above. The velocity of the ball as it leaves the kicker’s foot is \( 20 \, \text{m/s} \) at an angle of \( 37^\circ \) above the horizontal. The top of the fence is \( 2.5 \, \text{m} \) high. The ball hits nothing while in flight and air resistance is negligible.
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 golfer hits a shot to a green that is elevated \(2.80 \, \text{m}\) above the point where the ball is struck. The ball leaves the club at a speed of \(18.9 \, \text{m/s}\) at an angle of \(52.0^\circ\) above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
Water balloons are tossed from the roof of a building, all with the same speed but with different launch angles. Which one has the highest speed when it hits the ground? Ignore air resistance. Without using equations, explain your answer.
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?
a) 51.5 m
b) 25.3 m/s at 50.1° below the horizontal
c) 19.1 m
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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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