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| Derivation/Formula | Reasoning |
|---|---|
| \[v_i\sin\theta = 25(0.602)=15.0\,\text{m/s}\] | Resolve the launch speed into the vertical component using \(\sin37^\circ\approx0.602\). |
| \[\Delta y = v_i\sin\theta\,t – \tfrac12 g t^2\] | Use the vertical displacement equation with \(\Delta y = +5.0\,\text{m}\) and \(g=9.8\,\text{m/s}^2\). |
| \[4.9t^2-15.0t+5.0=0\] | Substitute numerical values and rearrange to standard quadratic form. |
| \[t=\frac{15.0\pm11.3}{9.8}\] | Apply the quadratic formula: \(t=\tfrac{-b\pm\sqrt{b^2-4ac}}{2a}\). |
| \[t=0.38\,\text{s}\quad\text{or}\quad t=2.69\,\text{s}\] | The smaller root is when the projectile is still rising; the larger root is the total flight time. |
| \[\boxed{t=2.69\,\text{s}}\] | Total time aloft. |
| Derivation/Formula | Reasoning |
|---|---|
| \[v_i\cos\theta = 25(0.799)=20.0\,\text{m/s}\] | Resolve the launch speed into the constant horizontal component using \(\cos37^\circ\approx0.799\). |
| \[\Delta x = v_i\cos\theta\;t\] | Horizontal displacement equals horizontal velocity times time; no horizontal acceleration. |
| \[\Delta x = 20.0\times2.69 = 5.38\times10^{1}\,\text{m}\] | Insert \(t=2.69\,\text{s}\) from part (a). |
| \[\boxed{\Delta x = 53.8\,\text{m}}\] | Horizontal range. |
| Derivation/Formula | Reasoning |
|---|---|
| \[v_{y\,(\text{top})}=0\,\text{m/s}\] | At the highest point the vertical component of velocity is momentarily zero for any projectile motion neglecting air resistance. |
| \[\boxed{0\,\text{m/s}}\] | Result. |
| Derivation/Formula | Reasoning |
|---|---|
| \[v_{y\,(\text{land})}=v_i\sin\theta – g t = 15.0 – 9.8(2.69) = -11.3\,\text{m/s}\] | Vertical velocity after \(t=2.69\,\text{s}\); negative sign indicates downward direction. |
| \[v_{x}=20.0\,\text{m/s}\] | Horizontal velocity is unchanged throughout flight. |
| \[v_x^2 = v_{y\,(\text{land})}^2 + (v_{x})^2 = (11.3)^2 + (20.0)^2 = 5.27\times10^{2}\] | Pythagorean addition of components for speed magnitude (labelled here as \(v_x\) per notation). |
| \[v_x = 22.9\,\text{m/s}\] | Square-root of previous result. |
| \[\boxed{v_x < v_i}\] | The landing speed \(22.9\,\text{m/s}\) is less than the launch speed \(25\,\text{m/s}\) because the projectile finishes 5 m higher, converting some kinetic energy into gravitational potential energy. |
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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.
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?
Projectiles 1 and 2 are launched from level ground at the same time and follow the trajectories shown in the figure. Which one of the projectiles, if either, returns to the ground first, and why?
Two objects are dropped from rest from the same height. Object \( A \) falls through a distance \( d_A \) during a time \( t \), and object \( B \) falls through a distance \( d_B \) during a time \( 2t \). If air resistance is negligible, what is the relationship between \( d_A \) and \( d_B \)?
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.
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?
An object is thrown straight upward at 64 m/s.
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.
An eagle is flying horizontally at \(6 \, \text{m/s}\) with a fish in its claws. It accidentally drops the fish.
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?
\(2.69\,\text{s}\)
\(53.8\,\text{m}\)
\(0\,\text{m/s}\)
\(v_x < v_i\)
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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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