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Derivation/Formula | Reasoning |
---|---|
\[ \Delta x = v_i \cos 30^{\circ}\, t \] | Horizontal motion: distance \( \Delta x = 45\,\text{m} \) equals horizontal speed times time. |
\[ t = \frac{\Delta x}{v_i \cos 30^{\circ}} \] | Solve previous relation algebraically for time \( t \). |
\[ t = \frac{45}{29.4\cos 30^{\circ}} \approx 1.77\,\text{s} \] | Insert the value of \( v_i \) found in part (b) to evaluate the time. |
\[ \boxed{t \approx 1.77\,\text{s}} \] | Time taken for the ball to reach the wall. |
Derivation/Formula | Reasoning |
---|---|
\[ y = y_0 + v_i \sin 30^{\circ}\, t – \tfrac12 g t^{2} \] | Vertical position of the ball; here \( y_0 = 1.5\,\text{m} \) and final \( y = 12.2\,\text{m} \). |
\[ 12.2 = 1.5 + v_i \sin 30^{\circ}\, t – 4.9 t^{2} \] | Substitute heights and \( g = 9.8\,\text{m/s}^2 \). |
\[ 10.7 = 0.5 v_i t – 4.9 t^{2} \] | Simplify numerical terms. |
\[ t = \frac{45}{v_i \cos 30^{\circ}} \] | Replace \( t \) using the horizontal relation from part (a). |
\[ 10.7 = 0.5 v_i \left(\frac{45}{v_i \cos 30^{\circ}}\right) – 4.9 \left(\frac{45}{v_i \cos 30^{\circ}}\right)^2 \] | Substitute the expression for \( t \) into the vertical equation. |
\[ v_i \approx 29.4\,\text{m/s} \] | Algebraic manipulation (no calculus) gives the speed; evaluate numerically. |
\[ \boxed{v_i \approx 29.4\,\text{m/s}} \] | Initial launch speed of the soccer ball. |
Derivation/Formula | Reasoning |
---|---|
\[ v_x = v_i \cos 30^{\circ} \] | Horizontal component remains constant. |
\[ v_x \approx 25.5\,\text{m/s} \] | Insert \( v_i = 29.4\,\text{m/s} \). |
\[ v_y = v_i \sin 30^{\circ} – g t \] | Vertical component at the wall. |
\[ v_y \approx -2.62\,\text{m/s} \] | Negative sign indicates downward motion. |
\[ v = \sqrt{v_x^{2} + v_y^{2}} \] | Magnitude of the resultant velocity vector. |
\[ v \approx 25.6\,\text{m/s} \] | Numerical magnitude. |
\[ \phi = \tan^{-1}\!\left(\frac{|v_y|}{v_x}\right) \approx 5.9^{\circ} \] | Angle the velocity makes with the horizontal; below because \( v_y < 0 \). |
\[ \boxed{v \approx 25.6\,\text{m/s\, at }5.9^{\circ}\text{ below horizontal}} \] | Complete velocity description as it clears the wall. |
Derivation/Formula | Reasoning |
---|---|
\[ t_2 = \frac{\Delta x}{v_i \cos 55.5^{\circ}} \] | Time to reach the wall with the higher angle. |
\[ t_2 \approx 2.70\,\text{s} \] | Evaluate using \( v_i = 29.4\,\text{m/s} \). |
\[ y_2 = y_0 + v_i \sin 55.5^{\circ} t_2 – \tfrac12 g t_2^{2} \] | Vertical position at the wall for the new angle. |
\[ y_2 \approx 31.3\,\text{m} \] | Numerical height above ground. |
\[ \boxed{y_2 > 9.0\,\text{m} \;\text{(clears wall)}} \] | The ball still passes far above the castle wall. |
Just ask: "Help me solve this problem."
One end of a spring is attached to a solid wall while the other end just reaches to the edge of a horizontal, frictionless tabletop, which is a distance \(h\) above the floor. A block of mass \(M\) is placed against the end of the spring and pushed toward the wall until the spring has been compressed a distance \(x\). The block is released and strikes the floor a horizontal distance \(D\) from the edge of the table. Air resistance is negligible. Derive expressions for the following quantities only in terms of \(M, x, D, h,\) and any constants.
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 train is moving to the right at \( 20 \) \( \text{m/s} \). A passenger on the train throws a ball horizontally to the left at \( 5 \) \( \text{m/s} \) (relative to the train).
A stone is thrown horizontally at \(8.0 \, \text{m/s}\) from a cliff \(80 \,\text{m}\) high. How far from the base of the cliff will the stone strike the ground?
A projectile is launched at a speed of \( 22 \) \( \text{m/s} \) at an angle of \( 60^{\circ} \) above the horizontal. It lands on a ramp that is \( 5 \) \( \text{m} \) lower than the launch height. How long does it take for the projectile to hit the ramp?
\(1.77\,\text{s}\)
\(29.4\,\text{m/s}\)
\(25.6\,\text{m/s at }5.9^{\circ}\text{ below horizontal}\)
\(31.3\,\text{m}>9.0\,\text{m (clears)}\)
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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] | 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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