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| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[v_{ix}=v_i\cos 30^{\circ},\;v_{iy}=v_i\sin 30^{\circ}\] | Resolve the initial speed \(v_i\) into horizontal and vertical components using the given launch angle \(30^{\circ}\). |
| 2 | \[t_1=\frac{40}{v_{ix}}\] | Time \(t_1\) to reach the wall is horizontal distance \(\Delta x_1=40\,\text{m}\) divided by horizontal speed \(v_{ix}\). |
| 3 | \[v_{wy}=v_{iy}-gt_1\] | Vertical speed at impact \(v_{wy}\) equals initial vertical speed minus gravitational drop over time \(t_1\). |
| 4 | \[y_w=v_{iy}t_1-\tfrac12gt_1^{2}\] | Vertical displacement gives the height \(y_w\) at which the projectile meets the wall. |
| 5 | \[t_2=\frac{55}{v_{ix}}\] | After the bounce the ball travels back \(40+15=55\,\text{m}\); horizontal speed magnitude is still \(v_{ix}\). |
| 6 | \[0=y_w+v_{wy}t_2-\tfrac12gt_2^{2}\] | Starting from height \(y_w\) with vertical speed \(v_{wy}\), it falls to ground (height 0) in time \(t_2\). |
| 7 | \[v_{ix}=\tfrac{\sqrt3}{2}v_i,\;v_{iy}=\tfrac12v_i\] | Insert the numeric cosine and sine of \(30^{\circ}\). |
| 8 | \[t_1=\frac{80}{\sqrt3\,v_i}\] | Substitute \(v_{ix}=\tfrac{\sqrt3}{2}v_i\) into the expression for \(t_1\). |
| 9 | \[y_w=\frac{40}{\sqrt3}-\frac{3200g}{3v_i^{2}}\] | Insert \(t_1\) from Step 8 into the height formula of Step 4 and simplify. |
| 10 | \[v_{wy}=\frac12v_i-\frac{80g}{\sqrt3\,v_i}\] | Substitute \(t_1\) into Step 3 for the vertical speed at the wall. |
| 11 | \[t_2=\frac{110}{\sqrt3\,v_i}\] | Use \(v_{ix}=\tfrac{\sqrt3}{2}v_i\) in the expression for \(t_2\). |
| 12 | \[0=\frac{95}{\sqrt3}-\frac{18050g}{3v_i^{2}}\] | Insert \(y_w,\;v_{wy},\;t_2\) into the vertical motion equation of Step 6 and combine like terms. |
| 13 | \[v_i^{2}=\frac{190g\sqrt3}{3}\] | Solve the algebraic equation of Step 12 for the square of the launch speed. |
| 14 | \[y_w=\frac{440}{19\sqrt3}\;\text{m}\] | Substitute \(v_i^{2}\) from Step 13 back into the height expression of Step 9. |
| 15 | \[\boxed{y_w\approx13.4\,\text{m}}\] | Numerical evaluation with \(g=9.8\,\text{m/s}^2\) gives the impact height on the wall. |
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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?
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A ball of mass \(m\) is released from rest at a distance \(h\) above a frictionless plane inclined at an angle of \(45^\circ\) to the horizontal as shown above. The ball bounces horizontally off the plane at point \(P_1\) with the same speed with which it struck the plane and strikes the plane again at point \(P_2\). In terms of \(g\) and \(h\), determine each of the following quantities:
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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?
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\(13.4\,\text{m}\)
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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