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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (a) Calculate the time taken for the ball to reach the wall. | ||
| 1 | Use horizontal motion equation:
\( x = v_0 \cos \theta \cdot t \) |
Relates horizontal distance, initial speed, angle, and time. |
| 2 | Use vertical motion equation:
\( y = y_0 + v_0 \sin \theta \cdot t – \tfrac{1}{2} g t^2 \) |
Relates vertical position, initial speed, angle, time, and gravity. |
| 3 | Solve for \( v_0 \) from horizontal equation:
\( v_0 = \dfrac{x}{t \cos \theta} \) |
Express \( v_0 \) in terms of known quantities and \( t \). |
| 4 | Substitute \( v_0 \) into vertical equation:
\( y = y_0 + \left( \dfrac{x}{t \cos \theta} \right) \sin \theta \cdot t – \tfrac{1}{2} g t^2 \) |
Eliminated \( v_0 \) to get an equation with only \( t \). |
| 5 | Substitute known values:
\( y = 12.2 \, \text{m} \) \( 12.2 = 1.5 + 45 \times 0.5774 – 4.9 t^2 \) |
Plugged in numerical values to simplify the equation. |
| 6 | Simplify the equation:
\( 12.2 = 1.5 + 25.983 – 4.9 t^2 \) |
Solved for \( t \), the time to reach the wall. |
| (b) Calculate the initial speed of the soccer ball. | ||
| 7 | Use \( v_0 = \dfrac{x}{t \cos \theta} \)
\( \cos 30^\circ = 0.8660 \) |
Calculated initial speed using time from part (a). |
| (c) Find the velocity of the soccer ball as it passes over the wall. | ||
| 8 | Compute horizontal velocity component:
\( v_x = v_0 \cos \theta = 29.38 \times 0.8660 \approx 25.45 \, \text{m/s} \) |
Horizontal velocity remains constant. |
| 9 | Compute vertical velocity at \( t = 1.77 \, \text{s} \):
\( v_y = v_0 \sin \theta – g t \) |
Calculated vertical velocity at the wall (negative indicates downward). |
| 10 | Compute magnitude of velocity:
\( v = \sqrt{v_x^2 + v_y^2} = \sqrt{(25.45)^2 + (-2.66)^2} \approx 25.58 \, \text{m/s} \) |
Found the speed as it passes over the wall. |
| 11 | Compute direction angle:
\( \phi = \arctan\left( \dfrac{v_y}{v_x} \right) = \arctan\left( \dfrac{-2.66}{25.45} \right) \approx -5.97^\circ \) |
Determined angle below horizontal. |
| (d) Determine if the ball clears the wall when fired at \( 55.5^\circ \). | ||
| 12 | Compute new time to reach wall:
\( \cos 55.5^\circ = 0.5657 \) |
Calculated time to reach wall with new angle. |
| 13 | Compute vertical position at \( t = 2.72 \, \text{s} \):
\( \sin 55.5^\circ = 0.8286 \) |
Found the height at the wall with new angle. |
| 14 | Compare height to wall height:
\( y_{\text{wall}} = 9.0 \, \text{m} \) |
Determined that the ball will still clear the wall. |
Just ask: "Help me solve this problem."
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On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 3.5 times as far as he would have on earth, given the same initial velocities on both planets. The ball is launched at a speed of 45 m/s at an angle of 29° above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet find:
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 cat chases a mouse across a 1.0 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 2.2 m from the edge of the table. When the cat slid off the table, what was its speed?
(a) The time taken for the ball to reach the wall is approximately \( 1.77 \, \text{s} \).
(b) The initial speed of the soccer ball is approximately \( 29.38 \, \text{m/s} \).
(c) As it passes over the wall, the ball’s velocity is approximately \( 25.58 \, \text{m/s} \) at an angle of \( 5.97^\circ \) below the horizontal.
(d) Yes, when fired at \( 55.5^\circ \), the ball will still clear the \( 9.0 \, \text{m} \) high wall, passing over it at a height of approximately \( 31.64 \, \text{m} \).
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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] |
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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