| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[v_i = 5.40~\text{m/s}\] \[y_0 = 105~\text{m}\] \[a = -9.80~\text{m/s}^2\] | Assign known quantities with upward positive, so gravity is \( a=-9.80~\text{m/s}^2 \). |
| 2 | \[y(t)=y_0 + v_i t + \tfrac{1}{2} a t^2\] | Use constant-acceleration position formula in one dimension. |
| 3 | \[0 = 105 + 5.40 t – 4.90 t^2\] | Set \( y(t)=0 \) for ground impact and substitute the given values. |
| 4 | \[4.90 t^2 – 5.40 t – 105 = 0\] | Multiply by \(-1\) to put the quadratic in standard form \( at^2+bt+c=0 \). |
| 5 | \[t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \]
\[= \frac{5.40 \pm \sqrt{(-5.40)^2 – 4(4.90)(-105)}}{2(4.90)}\] |
Apply quadratic formula with \( a=4.90,\; b=-5.40,\; c=-105 \). |
| 6 | \[t = \frac{5.40 \pm 45.69}{9.80}\] | Simplify the discriminant: \( (-5.40)^2 + 4(4.90)(105) = 2087.16 \) and take its square root \( \approx 45.69 \). |
| 7 | \[t = 5.21~\text{s}\] | Choose the positive root \( (5.40+45.69)/9.80 \) because time cannot be negative. |
| 8 | \[\boxed{t \approx 5.21~\text{s}}\] | Final time for the package to reach the ground. |
A Major Upgrade To Phy Is Coming Soon — Stay Tuned
We'll help clarify entire units in one hour or less — guaranteed.
A self paced course with videos, problems sets, and everything you need to get a 5. Trusted by over 15k students and over 200 schools.
An airplane accelerates down a runway at \( 10 \, \text{m/s}^2 \). It reaches a final velocity of \( 200 \, \text{m/s} \) until it finally lifts off the ground. Determine the distance traveled before takeoff.
On another planet, a ball is in free fall after being released from rest at time \( t = 0 \). A graph of the height of the ball above the planet’s surface as a function of time \( t \) is shown. The acceleration due to gravity on the planet is most nearly
The alarm at a fire station rings and a 79.34-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 4.20 m). Just before landing, his speed is 1.36 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?
An elevator of height \(h\) ascends with constant acceleration \(a\). When it crosses a platform, it has acquired a velocity \(u\). At this instant a bolt drops from the top of the elevator. Find the time for the bolt to hit the floor of the elevator. Give your answer in terms of \(h\), \(a\), and any constant.
A rocket is sent to shoot down an invading spacecraft that is hovering at an altitude of \( 1500 \, \text{m} \). The rocket is launched with an initial velocity of \( 180 \, \text{m/s} \). Find the following:
A body moving in the positive \( x \) direction passes the origin at time \( t = 0 \). Between \( t = 0 \) and \( t = 1 \, \text{second} \), the body has a constant speed of \( 24 \, \text{m/s} \). At \( t = 1 \, \text{second} \), the body is given a constant acceleration of \( 6 \, \text{m/s}^2 \) in the negative \( x \) direction. The position \( x \) of the body at \( t = 11 \, \text{seconds} \) is
You throw a ball straight upward. It leaves your hand at \( 20 \) \( \text{m/s} \) and slows at a steady rate until it stops at the peak. The ball then comes back down, speeding up steadily until it hits the ground with the same speed it left your hand. Draw the velocity vs. time graph or explain it in terms of functions.
You stand at the edge of a vertical cliff and throws a stone vertically upwards. The stone leaves your hand with a speed v = 8.0 m/s. The time between the stone leaving your hand and hitting the sea is 3.0 s. Assume air resistance is negligible. Calculate:
Which car controls directly allow the driver to cause acceleration?
\(5.21~\text{s}\)
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] |
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] |
One price to unlock most advanced version of Phy across all our tools.
per month
Billed Monthly. Cancel Anytime.
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
