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| Derivation / Formula | Reasoning |
|---|---|
| \[x(t) = v_i \cos\theta\, t\] | Horizontal motion is uniform (no horizontal acceleration), so position grows linearly with constant speed \(v_i\cos\theta\). |
| \[y(t) = v_i \sin\theta \, t – \tfrac{1}{2} g t^2\] | Vertical motion has initial upward speed \(v_i\sin\theta\) and constant downward acceleration \(g\), giving a parabola. |
| \[R=\frac{v_i^2\sin 2\theta}{g}\] | Level-ground range formula (derive by eliminating \(t_f\) from \(x=v_i\cos\theta\,t_f\) and \(0=v_i\sin\theta\,t_f-\tfrac{1}{2}gt_f^2\)). |
| \[\sin 2\theta=\frac{Rg}{v_i^2}=\frac{35\cdot 9.8}{20^2}=0.8575\] | Numerically \(2\theta\approx 59.0^\circ\) or \(121.0^\circ\), so \(\theta\approx 29.5^\circ\) (low) or \(\theta\approx 60.5^\circ\) (high). |
| \[h_{\max}=\frac{(v_i\sin\theta)^2}{2g}\] | Peak height test selects the physically relevant branch for \(y=5\,\text{m}\). |
| \[\begin{aligned} \theta&\approx 29.5^\circ:& h_{\max}&=\frac{(20\sin29.5^\circ)^2}{2\cdot 9.8}\approx 4.95\,\text{m}<5\\ \theta&\approx 60.5^\circ:& h_{\max}&=\frac{(20\sin60.5^\circ)^2}{2\cdot 9.8}\approx 15.46\,\text{m}>5 \end{aligned}\] | The low arc never reaches \(5\,\text{m}\); only the high arc can cross \(y=5\,\text{m}\) twice (upward and downward). |
| \[t=\frac{x}{v_i\cos\theta}\] | From \(x(t)=v_i\cos\theta\,t\), solve for time at a given horizontal position. |
| \[y(x)=v_i\sin\theta\Big(\frac{x}{v_i\cos\theta}\Big)-\frac{1}{2}g\Big(\frac{x}{v_i\cos\theta}\Big)^2\] | Substitute \(t\) into \(y(t)\) to express height directly in terms of \(x\). |
| \[y(x)=x\tan\theta-\frac{g\,x^2}{2v_i^{2}\cos^{2}\theta}\] | Algebraic simplification: \(\tan\theta=\frac{\sin\theta}{\cos\theta}\). |
| \[5=x\tan\theta-\frac{g\,x^2}{2v_i^{2}\cos^{2}\theta}\] | Impose the target height \(y=5\,\text{m}\) on the high-angle trajectory. |
| \[-\underbrace{\frac{g}{2v_i^{2}\cos^{2}\theta}}_{\displaystyle A}\,x^2+\underbrace{\tan\theta}_{\displaystyle B}\,x-\underbrace{5}_{\displaystyle C}=0\] | Identify quadratic coefficients \(a=-A,\; b=B,\; c=-5\). This makes the upcoming plug-in transparent. |
| \[\cos\theta\approx 0.4924,\quad \tan\theta\approx 1.7675,\quad A=\frac{9.8}{2\cdot 20^2\cos^2\theta}\approx 0.05052\] | Using \(\theta\approx 60.5^\circ\). Thus the quadratic is \(-0.05052\,x^2+1.7675\,x-5=0\). |
| \[\Delta=b^2-4ac=1.7675^2-4(-0.05052)(-5)\approx 2.114>0\] | Positive discriminant ⇒ two distinct horizontal positions reach \(y=5\,\text{m}\). |
| \[x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-1.7675\pm \sqrt{2.114}}{2(-0.05052)}\] | Quadratic formula with \(a=-0.05052,\; b=1.7675,\; c=-5\). |
| \[x\approx 3.10\,\text{m}\quad\text{or}\quad x\approx 31.88\,\text{m}\] | Two crossings of the \(5\,\text{m}\) level: once ascending, once descending. |
| \[\boxed{x=3.1\,\text{m},\;31.9\,\text{m}}\] | Final answer, rounded. |
Just ask: "Help me solve this problem."
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 \( H \)?
A ball is tossed straight up while the thrower is standing in a moving train car that is moving at a constant velocity. Neglecting air resistance, what is the path of the ball relative to the ground outside the train?
A circus cannon fires an acrobat into the air at an angle of \( 45^\circ \) above the horizontal, and the acrobat reaches a maximum height \( y \) above her original launch height. The cannon is now aimed so that it fires straight up, at an identical speed, into the air at an angle of \( 90^\circ \) to the horizontal. In terms of \( y \), what is the acrobat’s new maximum height?
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?
Seo-Jun throws a ball to her friend Zuri. The ball leaves Seo-Jun’s hand from a height \( h = 1.5 \) \( \text{m} \) above the ground with an initial speed \( \vec{v}_{s,0} = 12 \) \( \text{m/s} \) at an angle of \( \theta = 25^\circ \) with respect to the horizontal. Zuri catches the ball at a height of \( h = 1.5 \) \( \text{m} \) above the ground.
After catching the ball, Zuri throws it back to Seo-Jun. The ball leaves Zuri’s hand from a height \( h = 1.5 \) \( \text{m} \) above the ground. The ball is moving with a speed of \( 15 \) \( \text{m/s} \) when it reaches a maximum height of \( 5.8 \) \( \text{m} \) above the ground.
At what height \( h’ \) above the ground will the ball be when the return throw reaches Seo-Jun?
\(3.1 \text{ m}\)
\(31.9 \text{ m}\)
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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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