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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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A pistol is fired horizontally toward a target \(120 \,\text{m}\) away but at the same height. The bullet’s velocity is \(200 \, \text{m/s}\).
A gun can fire a bullet to height \( h \) when fired straight up. If the same gun is pointed at an angle of \( 45^\circ \) from the vertical, what is the new maximum height of the projectile?
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.
You must split an apple resting on top of you friend’s head from a distance of 27 m. When you aim directly at the apple, the arrow is horizontal. At what angle should you aim the arrow to hit the apple if the arrow travels at a speed of 35 m/s?
(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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