Step | Formula Derivation | Reasoning |
---|---|---|
1.1 | v_{x0} = v_0 \cos(\theta) | Calculate the horizontal component of the initial velocity. v_0 is the launch speed and \theta is the launch angle. |
1.2 | v_{x0} = 75.0 , \text{m/s} \times \cos(60.0^\circ) | Plug in v_0 = 75.0 , \text{m/s} and \theta = 60.0^\circ. |
1.3 | t = \frac{d}{v_{x0}} | Calculate the time t it takes for the rocket to reach the wall. d is the horizontal distance to the wall. |
1.4 | t = \frac{27.0 , \text{m}}{v_{x0}} | Plug in d = 27.0 , \text{m}. |
2.1 | v_{y0} = v_0 \sin(\theta) | Calculate the vertical component of the initial velocity. |
2.2 | v_{y0} = 75.0 , \text{m/s} \times \sin(60.0^\circ) | Plug in v_0 = 75.0 , \text{m/s} and \theta = 60.0^\circ. |
2.3 | y = v_{y0} t – \frac{1}{2} g t^2 | Use the kinematic equation for vertical motion. g is the acceleration due to gravity. |
2.4 | y = v_{y0} t – \frac{1}{2} (9.8 , \text{m/s}^2) t^2 | Plug in g = 9.8 , \text{m/s}^2. |
2.5 | y = v_{y0} \times \frac{27.0 , \text{m}}{v_{x0}} – \frac{1}{2} (9.8 , \text{m/s}^2) \left( \frac{27.0 , \text{m}}{v_{x0}} \right)^2 | Substitute t from step 1.4. |
3.1 | \text{Clearance} = y – 11.0 , \text{m} | Subtract the wall height from the rocket’s height to find the clearance. |
We will now calculate these steps.
The rocket clears the top of the wall by approximately 33.23 , \text{m}.
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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.
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 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 HA 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 HB below the center. Find the ratio НB/ НА.
A ball is kicked at a speed of v_0 an angle \theta above the horizontal. The ball travels 25 meters horizontally. If the ball is kicked at 2v_0 , what will the horizontal displacement be?
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.
33.23 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_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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