To solve the problem of finding the initial speed, v_0 , 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, R ) and it lands 0.05 meters below the center of the target (this is the vertical displacement, y ).
Here is the derivation using kinematic equations:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | y = \frac{1}{2} g t^2 | The vertical motion of the arrow is only affected by gravity, so we use the kinematic equation for displacement, where g is the acceleration due to gravity (approximately 9.81 m/s2), and t is the time of flight. |
| 2 | t = \sqrt{\frac{2y}{g}} | To find the time of flight, t , we rearrange the formula for displacement to solve for t . Plugging in y = 0.05 m, we can calculate the time. |
| 3 | t = \sqrt{\frac{2 \times 0.05}{9.81}} | Substitute the values into the formula to calculate t . |
| 4 | t \approx 0.1005 \text{ seconds} | Do the math to derive the approximate time value. |
| 5 | R = v_0 t | The horizontal motion is described by this equation, where R is the range (horizontal distance traveled) and v_0 is the initial horizontal velocity. |
| 6 | v_0 = \frac{R}{t} | Rearranging the equation allows us to solve for v_0 using the calculated t and given R , which is 20 meters. |
| 7 | v_0 = \frac{20}{0.1005} | Substitute t and R into the equation to find v_0 . |
| 8 | v_0 \approx 199.0 \text{ m/s} | 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.
Phy can also check your working. Just snap a picture!
A soccer ball is kicked horizontally off an 85-meter high cliff, at a speed of 34 m/s. What was the ball’s final speed when it hit the ground below?
A baseball is thrown at an angle of 25° relative to the ground at a speed of 23.0 m/s. The ball is caught 42.0 m from the thrower.
3 clay balls, labeled A, B, and C are launched from the same height at the same speed as shown below. A is launched at 30° above horizontal, B is launched horizontally, and C is launched 30° below the horizontal. They all hit the wall (before reaching the ground) in times tA, tB, and tC respectively. Rank these times from least to greatest.
On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 3.5 times as far as he would have on earth, given the same initial velocities on both planets. The ball is launched at a speed of 45 m/s at an angle of 29° above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet find:
A soccer ball with an initial height of 1.5 meter above the ground is launched at an angle of 30° above the horizontal. The cannonball travels a horizontal distance of 45 meters to a 9.0 meter high castle wall, and passes over 3.20 meters above the highest point of the wall. Assume air resistance is negligible.
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_1m_2}{r^2} |
| a = \frac{\Delta v}{\Delta t} | f = \mu N |
| R = \frac{v_i^2 \sin(2\theta)}{g} |
| Circular Motion | Energy |
|---|---|
| F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
| a_c = \frac{v^2}{r} | PE = mgh |
| KE_i + PE_i = KE_f + PE_f |
| Momentum | Torque and Rotations |
|---|---|
| p = m v | \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 |
|---|
| F = -k x |
| T = 2\pi \sqrt{\frac{l}{g}} |
| T = 2\pi \sqrt{\frac{m}{k}} |
| Constant | Description |
|---|---|
| g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
| G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
| \mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
| k | Spring constant, in \text{N/m} |
| M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
| M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
| M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| s (Displacement) | \text{meters (m)} |
| v (Velocity) | \text{meters per second (m/s)} |
| a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
| t (Time) | \text{seconds (s)} |
| m (Mass) | \text{kilograms (kg)} |
| Variable | Derived SI Unit |
|---|---|
| F (Force) | \text{newtons (N)} |
| E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
| P (Power) | \text{watts (W)} |
| p (Momentum) | \text{kilogram meters per second (kgm/s)} |
| \omega (Angular Velocity) | \text{radians per second (rad/s)} |
| \tau (Torque) | \text{newton meters (Nm)} |
| I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |
| f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \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}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | 10^{-12} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
The most advanced version of Phy. Currently 50% off, for early supporters.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Understand you mistakes quicker.
For GQs and FRQs, Phy provides brief feedback as to how you can improve your answer.
Aim to increase your understadning and average score with every attempt!
10 Free Credits To Get You Started
*Phy Pro members get unlimited credits
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
