(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 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.
An eagle is flying horizontally at \(6 \, \text{m/s}\) with a fish in its claws. It accidentally drops the fish.
A marble is thrown horizontally with a speed of \(15 \, \text{m/s}\) from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of \(65^\circ\) with the horizontal. From what height above the ground was the marble thrown?
The highest barrier that a projectile can clear is 16.2 m, when the projectile is launched at an angle of 22.0° above the horizontal. What is the projectile’s launch speed?
A officer fires a pistol horizontally toward a target \(120 \,\text{m}\) at a velocity of \(200 \, \text{m/s}\). If the officer aimed directly at the bull’s eye
A projectile is launched at an angle of \( 30^{\circ} \) and hits a vertical wall \( 40 \) \( \text{m} \) away. After bouncing back horizontally, it lands \( 15 \) \( \text{m} \) behind the launch point. How high up on the wall did the projectile strike?
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 batter hits a fly ball which leaves the bat \( 0.90 \) \( \text{m} \) above the ground at an angle of \( 61^\circ \) with an initial speed of \( 28 \) \( \text{m/s} \) heading toward centerfield. Ignore air resistance.
A javelin thrower standing at rest holds the center of the javelin behind her head, then accelerates it through a distance of \( 70 \, \text{cm} \) as she throws. She releases the \( 600 \, \text{g} \) javelin \( 2.0 \, \text{m} \) above the ground traveling at an angle of \( 30^\circ \) above the horizontal. In this throw, the javelin hits the ground \( 54 \, \text{m} \) away. Find the following:
A drinking fountain projects water at an initial angle of \( 50^ \circ \) above the horizontal, and the water reaches a maximum height of \( 0.150 \) \( \text{m} \) above the point of exit. Assume air resistance is negligible.
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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