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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( v = v_0 + at \) | Use the kinematic equation for vertical motion. Here, \(v\) is the final velocity at the max height (which is 0), \(v_0\) is the initial velocity, \(a\) is the acceleration (gravity, acting downward), and \(t\) is the time. |
| 2 | \( 0 = v_0 – g \times 0.586 \) | Set the final velocity \(v\) at the maximum height to 0, and solve for \(v_0\). The acceleration due to gravity \(g\) is \(9.8 \, \text{m/s}^2\). |
| 3 | \( v_0 = g \times 0.586 \) | Rearrange to solve for \(v_0\) |
| 4 | \( v_0 = 9.8 \times 0.586 \) | Substitute the values of \(g\) and \(t\) |
| 5 | \( v_0 = 5.74 \, \text{m/s} \) | Final answer for the initial speed. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( v^2 = v_0^2 + 2a \Delta x \) | Use the kinematic equation to relate velocity, acceleration, and displacement. Here, \(v\) is the final velocity at the max height (0), \(v_0\) is the initial velocity, \(a\) is the acceleration (gravity), and \(\Delta x\) is the displacement. |
| 2 | \( 0 = (5.74)^2 – 2 \cdot 9.8 \cdot \Delta x \) | Substitute \(v = 0\), \(v_0 = 5.74 \, \text{m/s}\), and \(a = 9.8 \, \text{m/s}^2\) |
| 3 | \( 2 \cdot 9.8 \cdot \Delta x = (5.74)^2 \) | Rearrange to solve for \(\Delta x\). |
| 4 | \( \Delta x = \frac{(5.74)^2}{2 \cdot 9.8} \) | Solve for \(\Delta x\). |
| 5 | \( \Delta x = 1.68 \, \text{m} \) | Compute the displacement \(\Delta x\). |
| 6 | \(\text{Max height} = 3.25 + 1.68 = 4.93 \, \text{m} \) | Since the ice cube was initially 3.25 m above the ground, add this to \(\Delta x\) to get the maximum height. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(\Delta y = v_0 t + \frac{1}{2} a t^2\) | Use the kinematic equation for vertical displacement, where \(\Delta y\) is the change in height, \(v_0\) is the initial velocity, \(a\) is the acceleration due to gravity, and \(t\) is the time. |
| 2 | \(\Delta y = 0\) at maximum height, then use \(y = 4.93 \, \text{m}\) | The maximum height the ice cube reaches is previously calculated as 4.93 m. |
| 3 | \( -4.93 \, \text{m} = – \frac{1}{2} \cdot 9.8 \cdot t^2\) | Solve for the time taken to fall from the maximum height to the ground. Note the displacement (\(\Delta y\)) is negative when falling down. |
| 4 | \(t = \sqrt{\frac{2 \cdot 4.93}{9.8}}\) | Substitute the values and solve for \(t\). |
| 5 | \(t = \sqrt{\frac{9.86}{9.8}} = \sqrt{1.01} \approx 1.005 \, \text{s}\) | Calculate the square root to find the fall time. |
| 6 | Total time = \(0.586 \, \text{s} (up) + 1.005\, \text{s} (down) \approx 1.59 \, \text{s}\) | Add the time going up (0.586 s) to the time coming down (1.005 s) to get the total time. |
| 7 | \( t_{\text{total}} = 1.59 \, \text{s} \) | Final time taken for the ice cube to reach the ground. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_y^2 = v_{max}^2 + 2 a \Delta y\) | Use the kinematic equation relating velocity, acceleration, and displacement, where \(v_y\) is the final velocity, \(v_{max}\) is the velocity at maximum height (0 m/s), \(a\) is the acceleration due to gravity, and \(\Delta y\) is the displacement (4.93 m). |
| 2 | \(v_y^2 = 0 + 2 \cdot 9.8 \cdot 4.93\) | Substitute the values into the equation. |
| 3 | \( v_y^2 = 96.508 \) | Calculate the right side of the equation. |
| 4 | \( v_y = \sqrt{96.508}\) | Take the square root to solve for \(v_y\). |
| 5 | \( v_y \approx 9.82 \, \text{m/s}\) | Final speed of the ice cube when it reaches the ground. |
| 6 | \( v_{\text{final}} \approx 9.82 \, \text{m/s} \) | The boxed final answer for the speed of the ice cube when it hits the ground. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_y^2 = v_{max}^2 + 2 a \Delta y\) | Use the kinematic equation relating velocity, acceleration, and displacement, where \(v_y\) is the final velocity (7.00 m/s), \(v_{max}\) is the velocity at maximum height (0 m/s), \(a\) is the acceleration due to gravity, and \(\Delta y\) is the displacement from the maximum height. |
| 2 | \( (7.00)^2 = 0 + 2 \cdot 9.8 \cdot \Delta y \) | Substitute the given speed and constants. |
| 3 | \( 49 = 19.6 \cdot \Delta y\) | Solve for \(\Delta y\) by isolating it on one side of the equation. |
| 4 | \( \Delta y = \frac{49}{19.6}\) | Rearrange to solve for \(\Delta y\). |
| 5 | \(\Delta y = 2.5 \, \text{m}\) | Final height change from the maximum height when the ice cube reaches a speed of 7.00 m/s downward. |
| 6 | \( \text{Height above ground} = 4.93 – 2.5 \) | Subtract the downwards displacement from the maximum height to get the current height above the ground. |
| 7 | \( \text{Height above ground} = 2.43 \, \text{m}\) | Final height of the ice cube above the ground when traveling at 7.00 m/s downward. |
| 8 | \( \text{Height} = 2.43 \, \text{m} \) | Boxed final answer. |
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A car increases its forward velocity uniformly from \(40 ~ \text{m/s}\) to \(80 ~ \text{m/s}\) while traveling a distance of \(200 ~ \text{m}\). What is its acceleration during this time?
A girl, standing still, tosses a ball vertically upwards. One second later, she tosses up another ball at the same velocity. The balls collide \( 0.5 \) \( \text{s} \) after the second ball is tossed. With what velocity were they tossed? The acceleration due to gravity is \( 9.8 \) \( \text{m/s}^2 \).
A rock is dropped from the top of a tall tower. Half a second later another rock, twice as massive as the first, is dropped. Ignoring air resistance and using ONLY simple kinematics (DO NOT use energy to explain this). Explain it like you would to a 5th grader and select the correct choice:
In which of the following cases does a car have a negative velocity and a positive acceleration? A car that is traveling in the:
A rollercoaster leaves the station at rest. Its speed increases steadily for \( 6 \) \( \text{s} \) as it heads down the first drop. The ride then levels out and it moves at a constant speed for \( 4 \) \( \text{s} \) before hitting the brakes and stopping in \( 3 \) \( \text{s} \). Draw the velocity vs. time graph or explain it in terms of functions.
Above is a graph of the \(distance\) vs. time for car moving along a road. According the graph, at which of the following times would the automobile have been accelerating positively?
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?
An object’s velocity \(v\) as a function of time \(t\) is given in the graph. Which of the following statements is true about the motion of the object?
A \(25 \, \text{g}\) steel ball is attached to the top of a \(24 \, \text{cm}\)-diameter vertical wheel of negligible mass. Starting from rest, the wheel accelerates at \(470 \, \text{rad/s}^2\). The ball is released after \(\frac{3}{4}\) of a revolution. How high does it go above the center of the wheel?
A spacecraft accelerates at a rate of \(20.0 \, \text{m/s}^2\).
Note answers may vary by \( \pm 0.2 \).
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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] |
Metric Prefixes
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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