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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."
A baseball is thrown at an angle of 25° relative to the ground at a speed of 23.0 m/s. The ball is caught 42.0 m from the thrower.
A ball is launched horizontally from a height. At the same time, another ball is dropped vertically from the same height. Which hits the ground first?
A cylindrical tank of water (height \( H \)) is punctured at a height \( h \) above the bottom. How far from the base of the tank will the water stream land (in terms of \( h \) and \( H \))? What must the value of \( h \) be such that the distance at which the stream lands will be equal to \( H \)?
3 clay balls, labeled A, B, and C are launched from the same height at the same speed as shown above. A is launched at \( 30^\circ \) above horizontal, B is launched horizontally, and C is launched \( 30^\circ \) below the horizontal. They all hit the wall (before reaching the ground) in times \( t_A \), \( t_B \), and \( t_C \) respectively. Rank these times from least to greatest.
A projectile is launched at \( 25 \) \( \text{m/s} \) at an angle of \( 45^\circ \). It lands on a slope \( 5 \) \( \text{m} \) below the launch height. On landing, it rebounds vertically with \( 80\% \) of its speed and falls straight down from there. Find the total time from launch to final impact at the base of the slope.
(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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