Projectile motion range is derived as follows:
Step | Formula Derivation | Reasoning |
---|---|---|
1.1 | v_{x} = v \cos(\theta) | Horizontal component of initial velocity, constant since no air resistance. |
1.2 | v_{y0} = v \sin(\theta) | Initial vertical component of the velocity. |
2.1 | t = \frac{2v_{y0}}{g} | Time of flight, derived from v_{y} = v_{y0} – gt and the fact that at maximum height, v_{y} = 0, hence t = \frac{2v_{y0}}{g} for a round trip. |
2.2 | t = \frac{2v \sin(\theta)}{g} | Substitute v_{y0} = v \sin(\theta). |
3.1 | R = v_{x} t | Range is the horizontal distance traveled. |
3.2 | R = v \cos(\theta) \times \frac{2v \sin(\theta)}{g} | Substitute v_{x} = v \cos(\theta) and t from step 2.2. |
3.3 | R = \frac{v^2 \sin(2\theta)}{g} | Simplify using trigonometric identity \sin(2\theta) = 2 \sin(\theta) \cos(\theta). |
Now we apply this to find the new range when the launch speed is doubled.
Step | Formula Derivation | Reasoning |
---|---|---|
4.1 | R' = \frac{(2v)^2 \sin(2\theta)}{g} | New range formula with doubled speed 2v. |
4.2 | R' = 4 \times \frac{v^2 \sin(2\theta)}{g} | Simplify to show the relationship with the original range. |
4.3 | R' = 4R | Substitute the original range R. |
Given the original range of 23 m, we will calculate the new range.
The new range, calculated from the derived formula with the launch speed doubled, is 92.0 , \text{m}.
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A javelin thrower standing at rest holds the center of the javelin behind her head, then accelerates it through a distance of 70 cm as she throws. She releases the 600 g javelin 2.0 m above the ground traveling at an angle of 30° above the horizontal. In this throw, the javelin hits the ground 54 m away. Find the following:
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.
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?
A golfer hits her ball in a high arcing shot. Air resistance is negligible. When the ball is at its highest point, which of the following is true?
A arrow is shot horizontally from a distance of 20 meters away. It lands .05 meters below the center of the target. If air resistance is negligible what was the initial speed of the arrow?
92
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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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