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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. |
Just ask: "Help me solve this problem."
A projectile is launched at \( 20 \) \( \text{m/s} \) and lands \( 35 \) \( \text{m} \) away on level ground. At what two horizontal positions is the projectile exactly \( 5.0 \) \( \text{m} \) above the ground?
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?
A major-league pitcher can throw a baseball in excess of \( 41.0 \, \text{m/s} \). If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is \( 17.0 \, \text{m} \) away from the point of release?
A projectile is launched at \( 25 \) \( \text{m/s} \) at an angle of \( 37^{\circ} \). It lands on a platform that is \( 5.0 \) \( \text{m} \) above the launch height.
A projectile is launched at angle \( \theta \) to the horizontal, with velocity \( v \), maximum vertical displacement \( s \), and angle \( \theta \) between \( 0^{\circ} \) and \( 45^{\circ} \). What will the maximum vertical displacement be if the projectile is now launched at an angle of \( 2 \theta \) from the horizontal with velocity \( v \)?
\( \boxed{h’ = 5.68\,m} \)
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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] |
General Metric Conversion Chart
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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