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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (a) Calculate the time taken for the ball to reach the wall. | ||
| 1 | Use horizontal motion equation:
\( x = v_0 \cos \theta \cdot t \) |
Relates horizontal distance, initial speed, angle, and time. |
| 2 | Use vertical motion equation:
\( y = y_0 + v_0 \sin \theta \cdot t – \tfrac{1}{2} g t^2 \) |
Relates vertical position, initial speed, angle, time, and gravity. |
| 3 | Solve for \( v_0 \) from horizontal equation:
\( v_0 = \dfrac{x}{t \cos \theta} \) |
Express \( v_0 \) in terms of known quantities and \( t \). |
| 4 | Substitute \( v_0 \) into vertical equation:
\( y = y_0 + \left( \dfrac{x}{t \cos \theta} \right) \sin \theta \cdot t – \tfrac{1}{2} g t^2 \) |
Eliminated \( v_0 \) to get an equation with only \( t \). |
| 5 | Substitute known values:
\( y = 12.2 \, \text{m} \) \( 12.2 = 1.5 + 45 \times 0.5774 – 4.9 t^2 \) |
Plugged in numerical values to simplify the equation. |
| 6 | Simplify the equation:
\( 12.2 = 1.5 + 25.983 – 4.9 t^2 \) |
Solved for \( t \), the time to reach the wall. |
| (b) Calculate the initial speed of the soccer ball. | ||
| 7 | Use \( v_0 = \dfrac{x}{t \cos \theta} \)
\( \cos 30^\circ = 0.8660 \) |
Calculated initial speed using time from part (a). |
| (c) Find the velocity of the soccer ball as it passes over the wall. | ||
| 8 | Compute horizontal velocity component:
\( v_x = v_0 \cos \theta = 29.38 \times 0.8660 \approx 25.45 \, \text{m/s} \) |
Horizontal velocity remains constant. |
| 9 | Compute vertical velocity at \( t = 1.77 \, \text{s} \):
\( v_y = v_0 \sin \theta – g t \) |
Calculated vertical velocity at the wall (negative indicates downward). |
| 10 | Compute magnitude of velocity:
\( v = \sqrt{v_x^2 + v_y^2} = \sqrt{(25.45)^2 + (-2.66)^2} \approx 25.58 \, \text{m/s} \) |
Found the speed as it passes over the wall. |
| 11 | Compute direction angle:
\( \phi = \arctan\left( \dfrac{v_y}{v_x} \right) = \arctan\left( \dfrac{-2.66}{25.45} \right) \approx -5.97^\circ \) |
Determined angle below horizontal. |
| (d) Determine if the ball clears the wall when fired at \( 55.5^\circ \). | ||
| 12 | Compute new time to reach wall:
\( \cos 55.5^\circ = 0.5657 \) |
Calculated time to reach wall with new angle. |
| 13 | Compute vertical position at \( t = 2.72 \, \text{s} \):
\( \sin 55.5^\circ = 0.8286 \) |
Found the height at the wall with new angle. |
| 14 | Compare height to wall height:
\( y_{\text{wall}} = 9.0 \, \text{m} \) |
Determined that the ball will still clear the wall. |
Just ask: "Help me solve this problem."
A javelin thrower standing at rest holds the center of the javelin behind her head, then accelerates it through a distance of \( 70 \, \text{cm} \) as she throws. She releases the \( 600 \, \text{g} \) javelin \( 2.0 \, \text{m} \) above the ground traveling at an angle of \( 30^\circ \) above the horizontal. In this throw, the javelin hits the ground \( 54 \, \text{m} \) away. Find the following:
A rifle is used to shoot a target twice, using identical cartridges. The first time, the rifle is aimed parallel to the ground and directly at the center of the bull’s-eye. The bullet strikes the target at a distance of \( H_A \) below the center, however. The second time, the rifle is similarly aimed, but from twice the distance from the target. This time the bullet strikes the target at a distance of \( H_B \) below the center. Find the ratio \( H_B / H_A \).
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 skier is accelerating down a \( 30.0^{\circ} \) hill at \( 3.80 \) \( \text{m/s}^2 \).
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?
(a) The time taken for the ball to reach the wall is approximately \( 1.77 \, \text{s} \).
(b) The initial speed of the soccer ball is approximately \( 29.38 \, \text{m/s} \).
(c) As it passes over the wall, the ball’s velocity is approximately \( 25.58 \, \text{m/s} \) at an angle of \( 5.97^\circ \) below the horizontal.
(d) Yes, when fired at \( 55.5^\circ \), the ball will still clear the \( 9.0 \, \text{m} \) high wall, passing over it at a height of approximately \( 31.64 \, \text{m} \).
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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] |
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] | |
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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