# Part (a): At what times is the object 20 m above the ground?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | y = v_0 t + \frac{1}{2} a t^2 | Use the kinematic equation for vertical displacement where y is the height, v_0 is the initial velocity, t is the time, and a is the acceleration. |
| 2 | 20 = 64 t – \frac{1}{2} \cdot 9.8 \cdot t^2 | Substitute y = 20 \text{ m}, v_0 = 64 \text{ m/s}, and a = -9.8 \text{ m/s}^2. Note the negative sign of acceleration due to gravity. |
| 3 | 4.9 t^2 – 64 t + 20 = 0 | Rearrange the equation into standard quadratic form a t^2 + b t + c = 0 where a = 4.9, b = -64, and c = 20. |
| 4 | t = \frac{64 \pm \sqrt{(-64)^2 – 4 \cdot 4.9 \cdot 20}}{2 \cdot 4.9} | Use the quadratic formula t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} to solve for t. |
| 5 | t_1 = \frac{64 + 60.88}{9.8} \approx 12.74 \text{ s}, \\ t_2 = \frac{64 – 60.88}{9.8} \approx 0.32 \text{ s} | Calculate the two possible values for t. The results are t_1 \approx 12.74 \text{ s} and t_2 \approx 0.32 \text{ s}. |
| 6 | t = 0.32 \text{ s}, 12.74 \text{ s} | Final answer: There are two times when the object is 20 m above the ground, at 0.32 \text{ s} and 12.74 \text{ s}. |
# Part (b): Why are there two answers for part (a)?
| Step | Explanation | Reasoning |
|---|---|---|
| 1 | Two Answers Explanation | There are two answers because the object passes 20 m twice: once while going up and once while coming back down. |
# Part (c): How long does it take to come back to the ground?
| Step | Derivation/Formula/Answer | Reasoning |
|---|---|---|
| 1 | y = v_0 t + \frac{1}{2} a t^2 | Use the kinematic equation for vertical displacement where y is zero as it returns to the ground. |
| 2 | 0 = 64 t – \frac{1}{2} \cdot 9.8 \cdot t^2 | Substitute y = 0 \text{ m}, v_0 = 64 \text{ m/s}, and a = -9.8 \text{ m/s}^2. |
| 3 | 0 = 64 t – 4.9 t^2 | Simplify the equation by calculating \frac{1}{2} \cdot 9.8 = 4.9. |
| 4 | 0 = t (64 – 4.9t) | Factor out t to solve for the time at which the object returns to the ground. |
| 5 | t = 0 \quad \text{or} \quad 64 = 4.9t \Rightarrow t = \frac{64}{4.9} \approx 13.06 \text{ s} | The first solution t = 0 corresponds to the initial throw. Solving 64 – 4.9t = 0 gives the time it takes to reach the ground. |
| 6 | t = 13.06 \text{ s} | Final answer: The object takes 13.06 \text{ s} to come back to the ground. |
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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} | |
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} |
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