Step | Formula / Calculation | Reasoning |
---|---|---|
1 | v_{x} = v_{0} \cos(\theta), v_{y} = v_{0} \sin(\theta) | Decomposing the initial velocity into horizontal (v_{x}) and vertical (v_{y}) components. |
2 | v_{x} = 97.5 \cos(50^\circ) \approx 62.68 , \text{m/s}, v_{y} = 97.5 \sin(50^\circ) \approx 74.67 , \text{m/s} | Calculating the horizontal and vertical components of the initial velocity. |
3 | y = y_{0} + v_{y} \cdot t – \frac{1}{2} g t^{2} | Kinematic equation for vertical motion, where y is the final height, y_{0} is the initial height, g is the acceleration due to gravity, and t is the time. |
4 | 0 = 732 + 74.67 \cdot t – \frac{1}{2} \cdot 9.81 \cdot t^{2} | Substituting values into the vertical motion equation to find the time when the package reaches the ground. |
5 | t \approx 11.80 s | Solving the quadratic equation for time, taking the positive root as the physical solution. |
6 | x = v_{x} \cdot t | Horizontal distance formula, where x is the distance and v_{x} is the constant horizontal velocity. |
7 | x = 62.68 \cdot 11.80 \approx 1379.29 m | Calculating the horizontal distance from the release point to where the package hits the earth. |
8 | v_{y \text{ final}} = v_{y} – g \cdot t | Calculating the final vertical velocity of the package just before impact. |
9 | v_{y \text{ final}} = 74.67 – 9.81 \cdot 11.80 \approx -36.05 m/s | Finding the final vertical component of velocity. |
10 | \theta_{\text{impact}} = \arctan\left(\frac{v_{y \text{ final}}}{v_{x}}\right) | Calculating the angle of the velocity vector with respect to the horizontal just before impact. |
11 | \theta_{\text{impact}} \approx -66.07^\circ | Evaluating the angle of impact. |
The distance along the ground from the point directly beneath the point of release to where the package hits the earth is approximately 1379.29 meters. The angle of the velocity vector of the package relative to the ground just before impact is approximately -66.07°.
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A person shoots a basket ball with a speed of 12 m/s at an angle of 35° above the horizontal. If the person is 2.4 m tall and the hoop is 3.05 m above the ground, how far back must the person stand in order to make the shot?
A ball of mass M is attached to a string of length L. It moves in a vertical circle and at the bottom the ball just clears the ground. The tension at the bottom of the path is 3 times the weight of the ball. Give all answers in terms of M, L, and g.
A circus cannon fires an acrobat into the air at an angle of 45° above the horizontal, and the acrobat reaches a maximum height y above her original launch height. The cannon is now aimed so that it fires straight up into the air at an angle of 90° to the horizontal. In terms of y, what is the acrobat’s new max height?
A diver springs upward from a diving board. At the instant she contacts the water her speed is 8.90 m/s, and her body is extended at an angle of 75.0° with respect to the horizontal surface of the water. At this instant her vertical displacement is -3.00 m, where downward is the negative direction. Determine her initial velocity, both magnitude and direction.
Suppose the water at the top of Niagara Falls has a horizontal speed of 2.7 m/s just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a 75° angle below the horizontal?
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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_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} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
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