0 attempts
0% avg
UBQ Credits
To solve the problem of finding the initial speed, [katex] v_0 [/katex], of an arrow shot horizontally, we can use the equations for projectile motion. The given parameters are that the arrow was shot from a horizontal distance of 20 meters (the range, [katex] R [/katex]) and it lands 0.05 meters below the center of the target (this is the vertical displacement, [katex] y [/katex]).
Here is the derivation using kinematic equations:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] y = \frac{1}{2} g t^2 [/katex] | The vertical motion of the arrow is only affected by gravity, so we use the kinematic equation for displacement, where [katex] g [/katex] is the acceleration due to gravity (approximately 9.81 m/s2), and [katex] t [/katex] is the time of flight. |
| 2 | [katex] t = \sqrt{\frac{2y}{g}} [/katex] | To find the time of flight, [katex] t [/katex], we rearrange the formula for displacement to solve for [katex] t [/katex]. Plugging in [katex]y = 0.05 [/katex] m, we can calculate the time. |
| 3 | [katex] t = \sqrt{\frac{2 \times 0.05}{9.81}} [/katex] | Substitute the values into the formula to calculate [katex] t [/katex]. |
| 4 | [katex] t \approx 0.1005 \text{ seconds} [/katex] | Do the math to derive the approximate time value. |
| 5 | [katex] R = v_0 t [/katex] | The horizontal motion is described by this equation, where [katex] R [/katex] is the range (horizontal distance traveled) and [katex]v_0 [/katex] is the initial horizontal velocity. |
| 6 | [katex] v_0 = \frac{R}{t} [/katex] | Rearranging the equation allows us to solve for [katex]v_0 [/katex] using the calculated [katex]t [/katex] and given [katex]R [/katex], which is 20 meters. |
| 7 | [katex] v_0 = \frac{20}{0.1005} [/katex] | Substitute [katex] t [/katex] and [katex] R [/katex] into the equation to find [katex]v_0 [/katex]. |
| 8 | [katex] v_0 \approx 199.0 \text{ m/s} [/katex] | Compute the result to get the initial speed of the arrow. |
This calculation shows that the initial speed of the arrow was approximately 199.0 meters per second.
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 block of mass M1 travels horizontally with a constant speed v0 on a plateau of height H until it comes to a cliff. A toboggan of mass M2 is positioned on level ground below the cliff. The center of the toboggan is a distance D from the base of the cliff.
A rock is thrown at an angle of \( 42^\circ \) above the horizontal at a speed of \( 14 \, \text{m/s} \). Determine how long it takes the rock to hit the ground.
Which of the following statements about the acceleration due to gravity is TRUE?
A ball of mass \(m\) is released from rest at a distance \(h\) above a frictionless plane inclined at an angle of \(45^\circ\) to the horizontal as shown above. The ball bounces horizontally off the plane at point \(P_1\) with the same speed with which it struck the plane and strikes the plane again at point \(P_2\). In terms of \(g\) and \(h\), determine each of the following quantities:
200 m/s
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] | |
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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
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.
