0 attempts
0% avg
UBQ Credits
Step | Derivation/Formula | Reasoning |
---|---|---|
1 | \( \Delta x = v_{i_x} t \) | Use the horizontal motion formula where \( \Delta x = 32 \, \text{m} \) and solve for \( t \). |
2 | \( v_{i_x} = v_i \cos(\theta) \) | Calculate the initial horizontal velocity using the initial velocity and the angle. |
3 | \( v_{i_x} = 20 \, \text{m/s} \cdot 0.80 = 16 \, \text{m/s} \) | Substitute \( v_i = 20 \, \text{m/s} \) and \( \cos(37^\circ) = 0.80 \) to find \( v_{i_x} \). |
4 | \( 32 \, \text{m} = 16 \, \text{m/s} \cdot t \) | Substitute \( \Delta x = 32 \, \text{m} \) and \( v_{i_x} = 16 \, \text{m/s} \) into the horizontal motion formula. |
5 | \( t = \frac{32 \, \text{m}}{16 \, \text{m/s}} \) | Rearrange to solve for \( t \). |
6 | \( t = 2 \, \text{s} \) | Calculate the value of \( t \). |
So, the time it takes for the ball to reach the plane of the fence is \( \boxed{2 \, \text{s}} \).
Step | Derivation/Formula | Reasoning |
---|---|---|
1 | \( y = v_{i_y} t + \frac{1}{2} a t^2 \) | Use the vertical motion equation to find the height \( y \) at \( t = 2 \, \text{s} \). |
2 | \( v_{i_y} = v_i \sin(\theta) \) | Calculate the initial vertical velocity using the initial velocity and the angle. |
3 | \( v_{i_y} = 20 \, \text{m/s} \cdot 0.60 = 12 \, \text{m/s} \) | Substitute \( v_i = 20 \, \text{m/s} \) and \( \sin(37^\circ) = 0.60 \) to find \( v_{i_y} \). |
4 | \( y = 12 \, \text{m/s} \cdot 2 \, \text{s} + \frac{1}{2} (-9.8 \, \text{m/s}^2) \cdot (2 \, \text{s})^2 \) | Substitute \( t = 2 \, \text{s} \), \( v_{i_y} = 12 \, \text{m/s} \), and \( a = -9.8 \, \text{m/s}^2 \) into the vertical motion formula. |
5 | \( y = 24 \, \text{m} – 19.6 \, \text{m} \) | Calculate the height \( y \). |
6 | \( y = 4.4 \, \text{m} \) | Determine the final value of \( y \) at \( t = 2 \, \text{s} \). |
7 | \( y_{\text{final}} – h_{\text{fence}} = 4.4 \, \text{m} – 2.5 \, \text{m} \) | Compare the calculated height with the fence height (\(2.5 \, \text{m}\)). |
8 | \( \Delta h = 1.9 \, \text{m} \) | Determine how far above the top of the fence the ball passes. |
So, the ball will pass \( \boxed{1.9 \, \text{m}} \) above the top of the fence.
Just ask: "Help me solve this problem."
Two balls are thrown off a building with the same speed, one straight up and one at a 45° angle. Which statement is true if air resistance can be ignored?
A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.0° above the horizontal. The rocket is fired toward an 11.0-m-high wall, which is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?
A ball is kicked horizontally off a 20 m tall cliff at a speed of 11 m/s. What is the final velocity of the ball right before it hits the ground?
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 into the air at an angle of \( 90^\circ \) to the horizontal. In terms of \( y \), what is the acrobat’s new maximum height?
The highest barrier that a projectile can clear is 16.2 m, when the projectile is launched at an angle of 22.0° above the horizontal. What is the projectile’s launch speed?
By continuing you (1) agree to our Terms of Sale and Terms of Use 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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
A quick explanation
Credits are used to grade your FRQs and GQs. Pro users get unlimited credits.
Submitting counts as 1 attempt.
Viewing answers or explanations count as a failed attempts.
Phy gives partial credit if needed
MCQs and GQs are are 1 point each. FRQs will state points for each part.
Phy customizes problem explanations based on what you struggle with. Just hit the explanation button to see.
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.