| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (a) How fast is the rocket traveling when the engine cuts off? | ||
| 1 | Use the kinematic equation:
\( v^2 = u^2 + 2 a s \) |
Relates final velocity, initial velocity, acceleration, and distance. |
| 2 | Substitute values:
\( v^2 = 0 + 2 \times 12.0 \times 1,000 \) |
Calculated \( v^2 \) using given values. |
| 3 | Solve for \( v \):
\( v = \sqrt{24,000} \approx 154.92 \, \text{m/s} \) |
Found the rocket’s speed at engine cutoff. |
| (b) What maximum height relative to the ground does the rocket reach? | ||
| 4 | After engine cutoff, use \( v^2 = u^2 + 2 a s \) with \( v = 0 \, \text{m/s} \), \( u = 154.92 \, \text{m/s} \), \( a = -9.8 \, \text{m/s}^2 \):
\( 0 = (154.92)^2 + 2 (-9.8) s \) |
Used kinematic equation for upward motion until velocity is zero. |
| 5 | Solve for \( s \):
\( s = \dfrac{(154.92)^2}{2 \times 9.8} \) |
Calculated additional height after engine cutoff. |
| 6 | Total maximum height:
\( h_{\text{total}} = 1,000 + 1,224.49 = 2,224.49 \, \text{m} \) |
Added initial altitude to additional height for total height. |
| (c) Velocity just before the rocket hits the earth. | ||
| 7 | Use \( v^2 = u^2 + 2 a s \) for free fall from maximum height with \( u = 0 \, \text{m/s} \), \( a = 9.8 \, \text{m/s}^2 \), \( s = 2,224.49 \, \text{m} \):
\( v^2 = 0 + 2 \times 9.8 \times 2,224.49 \) |
Calculated final velocity during descent. |
| 8 | Compute \( v^2 \):
\( v^2 = 43,598.00 \, \text{m}^2/\text{s}^2 \) |
Found velocity just before impact. |
| (d) Total amount of time the rocket was in the air. | ||
| 9 | **Time during powered ascent:**
Use \( s = u t + \dfrac{1}{2} a t^2 \) with \( s = 1,000 \, \text{m} \), \( u = 0 \, \text{m/s} \), \( a = 12.0 \, \text{m/s}^2 \): |
Calculated time for powered ascent. |
| 10 | Solve for \( t \):
\( t^2 = \dfrac{1,000}{6.0} \approx 166.67 \) |
Found time during powered ascent. |
| 11 | **Time during coasting ascent:**
Use \( v = u + a t \) with \( v = 0 \, \text{m/s} \), \( u = 154.92 \, \text{m/s} \), \( a = -9.8 \, \text{m/s}^2 \): |
Calculated time from engine cutoff to maximum height. |
| 12 | Solve for \( t \):
\( t = \dfrac{154.92}{9.8} \approx 15.81 \, \text{s} \) |
Found time during coasting ascent. |
| 13 | **Time during free fall descent:**
Use \( s = \dfrac{1}{2} a t^2 \) with \( s = 2,224.49 \, \text{m} \), \( a = 9.8 \, \text{m/s}^2 \): |
Calculated time for descent back to earth. |
| 14 | Solve for \( t \):
\( t^2 = \dfrac{2,224.49}{4.9} \approx 454.9996 \) |
Found time during free fall descent. |
| 15 | **Total time in the air:**
\( t_{\text{total}} = t_{\text{ascent}} + t_{\text{coasting}} + t_{\text{descent}} \) |
Summed all time intervals for total flight time. |
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.
A stone is thrown vertically upward with a speed of \( 24.0 \) \( \text{m/s} \).
A driver is driving at \( 40 \, \text{m/s} \) when the light turns red in front of her. It takes the driver \( 0.9 \, \text{s} \) to react and hit the brakes. After this, the car slows with an acceleration of \( 3.5 \, \text{m/s}^2 \). What is the total distance traveled by the car?
Which pair of quantities will always have the same magnitude if motion is in a straight line and in one direction?
A teacher walks the following path in \( 10 \) \( \text{s} \): \( 2 \) \( \text{m} \) south, \( 4 \) \( \text{m} \) east, \( 2 \) \( \text{m} \) north, \( 4 \) \( \text{m} \) west. What is the teacher’s average velocity?
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:
An object is projected vertically upward from ground level. It rises to a maximum height \( H \). If air resistance is negligible, which of the following must be true for the object when it is at a height \( H/2 \) ?
A rubber ball bounces on the ground. After each bounce, the ball reaches one-half the height of the bounce before it. If the time the ball was in the air between the first and second bounce was 1 second. What would be the time between the second and third bounce?
Traveling at a speed of 15.9 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.659. What is the speed of the automobile after 1.59 s have elapsed? Ignore the effects of air resistance.
A ball is dropped from a window \(10 \, \) above the sidewalk. Determine the time it takes for the ball to fall to the sidewalk.
A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard \( 3.4 \) \( \text{s} \) later. If the speed of sound is \( 340 \) \( \text{m/s} \), how high is the cliff?
Note: Answers may be off by \( \pm 0.2 \).
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.
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
