# Part (a): What is the difference in the time the balls spend in the air?
Step | Derivation/Formula | Reasoning |
---|---|---|
1 | y = v_0 t + \frac{1}{2}gt^2 | Use the kinematic equation for vertical motion to find the time of flight of each ball. |
2 | 19.6 = 0 + 14.7t + \frac{1}{2}(9.8)t^2 | For the ball thrown downward: v_0 = -14.7 \, \text{m/s} , g = 9.8 \, \text{m/s}^2 |
3 | 19.6 = 14.7t + 4.9t^2 \Rightarrow 0 = 4.9t^2 + 14.7t – 19.6 | Solve the quadratic equation using the quadratic formula. |
4 | t = \frac{-14.7 \pm \sqrt{(14.7)^2 – 4(4.9)(-19.6)}}{2(4.9)} | The positive root is taken since time cannot be negative. |
5 | t = 1 \, \text{s} | Solving the quadratic equation gives t = 1 \, \text{s} . |
6 | y = v_0 t + \frac{1}{2}gt^2 | Now for the ball thrown upward: initial speed v_0 = 14.7 \, \text{m/s}</td> </tr> <tr> <td>7</td> <td>0 = 14.7t – \frac{1}{2}(9.8)t^2 \Rightarrow t(2.9 – 2.9t) = 0 \Rightarrow t = 3 \, \text{s}</td> <td>Solving for \( t when the ball reaches the top of its trajectory. |
8 | \text{Total time in air} = (2 \times 1 + 3)\ s = 5 \, \text{.} | The time the second ball spends in the air is double the ascent time plus the fall time. |
9 | Difference in time = 2 \, \text{s} – 1\,\text{s} = 1s | The difference in the time the balls spend in the air. |
# Part (b): What is the velocity of each ball as it strikes the ground?
Step | Derivation/Formula | Reasoning |
---|---|---|
1 | v = v_0 + gt | Use the first kinematic equation to find the final velocity. |
2 | v = -14.7 + 9.8 \times 1 = -24.5 \, \text{m/s} | For the ball thrown downward: add gravitational acceleration effect. |
3 | v = 14.7 – 9.8 \times 3 = -14.7 \, \text{m/s} | For the ball thrown upward: subtract the effect of gravity over fall time. |
# Part (c): How far apart are the balls 0.800 s after they are thrown?
Step | Derivation/Formula | Reasoning |
---|---|---|
1 | d_{downward} = v_0 t + \frac{1}{2}gt^2 | Use kinematic equation for the ball thrown downward. |
2 | d_{downward} = 14.7 \times 0.800 + \frac{1}{2} \times 9.8 \times 0.800^2 = 15.68 \, \text{m} | Calculate distance traveled by the downward-thrown ball in 0.800s. |
3 | d_{upward} = v_0 t – \frac{1}{2}gt^2 | Use kinematic equation for the ball thrown upward. |
4 | d_{upward} = 14.7 \times 0.800 – \frac{1}{2} \times 9.8 \times 0.800^2 = 6.864 \, \text{m} | Calculate distance traveled by the upward-thrown ball in 0.800s. |
5 | 20 \, \text{m} | Distance between the balls after 0.800 s is 15.68\, \text{m} + 6.864\, \text{m} = 22.544 |
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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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