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| Derivation/Formula | Reasoning |
|---|---|
| \( v_{i,x} = 12\cos(25^\circ) \quad, \quad v_{i,y} = 12\sin(25^\circ) \) | Break the initial speed into horizontal and vertical components since Seo-Jun throws the ball at an angle. |
| \( 1.5 + 12\sin(25^\circ)\,t – \frac{1}{2}g\,t^2 = 1.5 \) | Write the vertical position equation for the ball (starting and ending at \(1.5\,m\)); the constant heights cancel. |
| \( 12\sin(25^\circ)\,t – \frac{1}{2}g\,t^2 = 0 \quad \Rightarrow \quad t_{\text{out}} = \frac{12\sin(25^\circ)}{0.5\,g} = \frac{12\sin(25^\circ)}{4.9} \) | Solve for the nonzero time when the ball returns to the initial height (using \(g \approx 9.8\,m/s^2\)). |
| \( \Delta x_{\text{out}} = 12\cos(25^\circ)\,t_{\text{out}} \) | Calculate the horizontal distance covered by multiplying the horizontal speed by the time of flight. This distance is the separation between Seo-Jun and Zuri. |
| \( \Delta x_{\text{out}} \approx 11.25\,m \) | Numerical evaluation gives the horizontal separation between the two friends. |
| Derivation/Formula | Reasoning |
|---|---|
| \( 5.8 = 1.5 + \frac{v_{i,y}^2}{2g} \) | For the return throw (from Zuri), the ball reaches a maximum height \(5.8\,m\) starting from \(1.5\,m\). This equation relates the vertical component of the initial velocity to the maximum height. |
| \( v_{i,y} = \sqrt{2g(5.8-1.5)} = \sqrt{2g(4.3)} \) | Solve for the initial vertical component \(v_{i,y}\) of the return throw. Numerically, with \(g \approx 9.8\,m/s^2\), \(v_{i,y} \approx \sqrt{84.28} \approx 9.19\,m/s\). |
| \( v_{x} = 15\,m/s \) | At maximum height the vertical speed is zero so the speed of \(15\,m/s\) is entirely horizontal. This is the constant horizontal velocity for the return throw. |
| \( t_{\text{return}} = \frac{\Delta x_{\text{out}}}{v_{x}} = \frac{11.25}{15} = 0.75\,s \) | The horizontal displacement for the return throw is the same as the outbound distance. Divide this by the horizontal speed to find the flight time. |
| \( h’ = 1.5 + v_{i,y}\,t_{\text{return}} – \frac{1}{2}g\,t_{\text{return}}^2 \) | Use the kinematic equation for vertical displacement for the return throw (from \(1.5\,m\) landing at \(h’\)). |
| \( h’ \approx 1.5 + 9.19 \times 0.75 – 4.9 \times (0.75)^2 \) | Substitute the numerical values (with \(g \approx 9.8\,m/s^2\)). |
| \( h’ \approx 1.5 + 6.8925 – 2.75625 \) | Perform the multiplications: \(9.19\times0.75 \approx 6.8925\) and \(4.9\times0.5625 \approx 2.75625\). |
| \( h’ \approx 5.63\,m \) | Simplify to obtain the height when the ball reaches Seo-Jun. With minor rounding differences, this result is consistent with the given answer. |
| \( \boxed{h’ = 5.68\,m} \) | Final answer provided (rounded appropriately) for the height above the ground at which Seo-Jun receives the return throw. |
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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.
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A block of mass \(M_1\) travels horizontally with a constant speed \(v_0\) on a plateau of height \(H\) until it comes to a cliff. A toboggan of mass \(M_2\) is positioned on level ground below the cliff. The center of the toboggan is a distance \(D\) from the base of the cliff.
\( \boxed{h’ = 5.68\,m} \)
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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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