0 attempts
0% avg
UBQ Credits
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| Initial Information | ||
| 0 | – Airplane speed: \( v_p = 97.5 \, \text{m/s} \) – Climb angle: \( \theta = 50.0^\circ \) – Altitude at release: \( h = 732 \, \text{m} \) – Acceleration due to gravity: \( g = 9.8 \, \text{m/s}^2 \) |
Gathered all given data for the problem. |
| 1 | Resolve initial velocity into components:
– Horizontal component: – Vertical component: |
Calculated initial horizontal (\( v_{0x} \)) and vertical (\( v_{0y} \)) velocities of the package. |
| 2 | Set up the vertical motion equation to find time \( t \):
\( y = y_0 + v_{0y} t – \dfrac{1}{2} g t^2 \) |
Established the equation for vertical motion to solve for time. |
| 3 | Rearrange the equation into a quadratic form:
\( -\dfrac{1}{2} g t^2 + v_{0y} t + h = 0 \) |
Prepared the quadratic equation to solve for \( t \). |
| 4 | Calculate the discriminant \( D \):
\( D = b^2 – 4ac \) |
Computed the discriminant to use in the quadratic formula. |
| 5 | Solve for time \( t \) using the quadratic formula:
\( t = \dfrac{-b \pm \sqrt{D}}{2a} \) |
Found the time of flight \( t \) for the package. |
| 6 | (a) Calculate the horizontal distance traveled:
\( x = v_{0x} t \) |
Determined the horizontal distance from the release point to impact. |
| 7 | Calculate the final vertical velocity \( v_y \):
\( v_y = v_{0y} – g t \) |
Computed the vertical component of velocity just before impact. |
| 8 | Calculate the magnitude of the final velocity \( v \):
\( v = \sqrt{v_{x}^2 + v_{y}^2} \) |
Found the speed of the package just before impact. |
| 9 | (b) Determine the angle \( \phi \) of the velocity vector:
\( \tan \phi = \dfrac{|v_y|}{v_x} = \dfrac{141.28}{62.73} \approx 2.252 \) |
Calculated the angle of the velocity vector relative to the ground. |
Just ask: "Help me solve this problem."
A car accelerates from rest with an acceleration of \( 3.5 \, \text{m/s}^2 \) for \( 10 \, \text{s} \). After this, it continues at a constant speed for an unknown amount of time. The driver notices a ramp \( 50 \, \text{m} \) ahead and takes \( 0.6 \, \text{s} \) to react. After reacting, the driver hits the brakes, which slow the car with an acceleration of \( 7.2 \, \text{m/s}^2 \). Unfortunately, the driver does not stop in time and goes off the \( 3 \, \text{m} \) high ramp that is angled at \( 27^\circ \).
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?
A marble is thrown horizontally with a speed of \(15 \, \text{m/s}\) from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of \(65^\circ\) with the horizontal. From what height above the ground was the marble thrown?
A rifle is used to shoot a target twice, using identical cartridges. The first time, the rifle is aimed parallel to the ground and directly at the center of the bull’s-eye. The bullet strikes the target at a distance of \( H_A \) below the center, however. The second time, the rifle is similarly aimed, but from twice the distance from the target. This time the bullet strikes the target at a distance of \( H_B \) below the center. Find the ratio \( H_B / H_A \).
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) \( 1379.29 \, \text{m} \)
(b) \( 66.1^\circ \) below the horizontal.
By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
We crafted THE Ultimate A.P Physics 1 Program so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the 2026 AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed
