| Derivation / Formula | Reasoning |
|---|---|
| \[v_x = 15\,\text{m/s}\] | Horizontal speed stays constant; therefore \(v_x = v_i = 15\,\text{m/s}\). |
| \[v_y = v_x \tan 65^\circ\] | At impact the velocity makes \(65^\circ\) with horizontal, so \(\tan 65^\circ = v_y / v_x\). |
| \[v_y = (15\,\text{m/s})\tan 65^\circ \approx 32.2\,\text{m/s}\] | Calculate the vertical component: \(v_y = 15\tan 65^\circ \approx 32.2\,\text{m/s}\) downward. |
| \[v_y^2 = v_{i,y}^2 + 2g\Delta y\] | Use the kinematic relation for vertical motion with constant acceleration \(g\). |
| \[\Delta y = \frac{v_y^2}{2g}\] | Initial vertical speed is zero \((v_{i,y}=0)\), so rearrange to find \(\Delta y\). |
| \[\Delta y = \frac{(32.2\,\text{m/s})^2}{2(9.8\,\text{m/s}^2)} \approx 5.28 \times 10^{1}\,\text{m}\] | Substitute \(v_y\) and \(g = 9.8\,\text{m/s}^2\) to obtain the height. |
| \[\boxed{h \approx 52.8\,\text{m}}\] | Express the height from which the marble was thrown. |
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In archery, should the arrow be aimed directly at the target? How should your angle of aim depend on the distance to the target? Explain without using equations.
Water balloons are tossed from the roof of a building, all with the same speed but with different launch angles. Which one has the highest speed when it hits the ground? Ignore air resistance. Without using equations, explain your answer.
A javelin thrower standing at rest holds the center of the javelin behind her head, then accelerates it through a distance of \( 70 \, \text{cm} \) as she throws. She releases the \( 600 \, \text{g} \) javelin \( 2.0 \, \text{m} \) above the ground traveling at an angle of \( 30^\circ \) above the horizontal. In this throw, the javelin hits the ground \( 54 \, \text{m} \) away. Find the following:
A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.0° above the horizontal. The rocket is fired toward an 11.0-m-high wall, which is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?
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?
Consider a ball thrown up from the surface of the earth into the air at an angle of \( 30^\circ \) above the horizontal. Air resistance is negligible. The ball’s acceleration just after release is most nearly
A ball is kicked at a speed of \( v_0 \) at an angle \( \theta \) above the horizontal. The ball travels 25 meters horizontally. If the ball is kicked at \( 2v_0 \), what will the horizontal displacement be?
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?
A batter hits a fly ball which leaves the bat \( 0.90 \) \( \text{m} \) above the ground at an angle of \( 61^\circ \) with an initial speed of \( 28 \) \( \text{m/s} \) heading toward centerfield. Ignore air resistance.
Measurements made in 1910 indicate that the common flea is an impressive jumper, given its size. Assume that a flea’s initial speed is 2.1 m/s, and that it leaps at an angle of 21° with respect to the horizontal. The jump lasts 0.16 s.
\(52.8\,\text{m}\)
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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] | |
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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