| Derivation/Formula | Reasoning |
|---|---|
| \( v_{i,x} = 12\cos(25^\circ) \quad, \quad v_{i,y} = 12\sin(25^\circ) \) | Break the initial speed into horizontal and vertical components since Seo-Jun throws the ball at an angle. |
| \( 1.5 + 12\sin(25^\circ)\,t – \frac{1}{2}g\,t^2 = 1.5 \) | Write the vertical position equation for the ball (starting and ending at \(1.5\,m\)); the constant heights cancel. |
| \( 12\sin(25^\circ)\,t – \frac{1}{2}g\,t^2 = 0 \quad \Rightarrow \quad t_{\text{out}} = \frac{12\sin(25^\circ)}{0.5\,g} = \frac{12\sin(25^\circ)}{4.9} \) | Solve for the nonzero time when the ball returns to the initial height (using \(g \approx 9.8\,m/s^2\)). |
| \( \Delta x_{\text{out}} = 12\cos(25^\circ)\,t_{\text{out}} \) | Calculate the horizontal distance covered by multiplying the horizontal speed by the time of flight. This distance is the separation between Seo-Jun and Zuri. |
| \( \Delta x_{\text{out}} \approx 11.25\,m \) | Numerical evaluation gives the horizontal separation between the two friends. |
| Derivation/Formula | Reasoning |
|---|---|
| \( 5.8 = 1.5 + \frac{v_{i,y}^2}{2g} \) | For the return throw (from Zuri), the ball reaches a maximum height \(5.8\,m\) starting from \(1.5\,m\). This equation relates the vertical component of the initial velocity to the maximum height. |
| \( v_{i,y} = \sqrt{2g(5.8-1.5)} = \sqrt{2g(4.3)} \) | Solve for the initial vertical component \(v_{i,y}\) of the return throw. Numerically, with \(g \approx 9.8\,m/s^2\), \(v_{i,y} \approx \sqrt{84.28} \approx 9.19\,m/s\). |
| \( v_{x} = 15\,m/s \) | At maximum height the vertical speed is zero so the speed of \(15\,m/s\) is entirely horizontal. This is the constant horizontal velocity for the return throw. |
| \( t_{\text{return}} = \frac{\Delta x_{\text{out}}}{v_{x}} = \frac{11.25}{15} = 0.75\,s \) | The horizontal displacement for the return throw is the same as the outbound distance. Divide this by the horizontal speed to find the flight time. |
| \( h’ = 1.5 + v_{i,y}\,t_{\text{return}} – \frac{1}{2}g\,t_{\text{return}}^2 \) | Use the kinematic equation for vertical displacement for the return throw (from \(1.5\,m\) landing at \(h’\)). |
| \( h’ \approx 1.5 + 9.19 \times 0.75 – 4.9 \times (0.75)^2 \) | Substitute the numerical values (with \(g \approx 9.8\,m/s^2\)). |
| \( h’ \approx 1.5 + 6.8925 – 2.75625 \) | Perform the multiplications: \(9.19\times0.75 \approx 6.8925\) and \(4.9\times0.5625 \approx 2.75625\). |
| \( h’ \approx 5.63\,m \) | Simplify to obtain the height when the ball reaches Seo-Jun. With minor rounding differences, this result is consistent with the given answer. |
| \( \boxed{h’ = 5.68\,m} \) | Final answer provided (rounded appropriately) for the height above the ground at which Seo-Jun receives the return throw. |
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If a baseball pitch leaves the pitcher’s hand horizontally at a velocity of \( 150 \) \( \text{km/h} \), by what \( \% \) will the pull of gravity change the magnitude of the velocity when the ball reaches the batter, \( 18 \) \( \text{m} \) away? For this estimate, ignore air resistance and spin on the ball.
A textbook is launched up with a speed of 20 m/s, at an angle of 36°, from a 12 m high roof.
Which of the following statements about the acceleration due to gravity is TRUE?
Three identical rocks are launched with identical speeds from the top of a platform of height \( h_0 \).
Which of the following correctly relates the magnitude \( v_y \) of the vertical component of the velocity of each rock immediately before it hits the ground?
A train is moving to the right at \( 20 \) \( \text{m/s} \). A passenger on the train throws a ball horizontally to the left at \( 5 \) \( \text{m/s} \) (relative to the train).
A ball is launched and lands \( 20 \) \( \text{m} \) away below the launch point \( 2.5 \) \( \text{s} \) later. The maximum height reached is \( 8.0 \) \( \text{m} \). What was the original launch velocity?
Two balls are launched at the same speed. Ball A is launched at an angle of \( 45^{\circ} \) and Ball B is launched at an angle of \( 60^{\circ} \). Which one reaches a higher point?
A golfer hits a shot to a green that is elevated \(2.80 \, \text{m}\) above the point where the ball is struck. The ball leaves the club at a speed of \(18.9 \, \text{m/s}\) at an angle of \(52.0^\circ\) above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
A rocket-powered hockey puck has a thrust of \(4.40 \, \text{N}\) and a total mass of \(1.00 \, \text{kg}\). It is released from rest on a frictionless table, \(2.10 \, \text{m}\) from the edge of a \(2.10 \, \text{m}\) drop. The front of the rocket is pointed directly toward the edge. Assuming that the thrust of the rocket is present for the entire time of travel, how far does the puck land from the base of the table?
During projectile motion (neglecting air resistance), what is the vertical acceleration at the highest point, assuming the initial velocity is upwards in the positive direction?
\( \boxed{h’ = 5.68\,m} \)
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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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