Projectile Motion and Vectors Review

1. Independence of Horizontal and Vertical Motion

Projectile motion can be analyzed by separating it into independent horizontal and vertical components. Air resistance is typically ignored in these problems.

2. Vertical Motion Under Gravity

• The vertical component of velocity ( $v_y$ ) is affected by gravitational acceleration ( $g$ ), which acts downwards. Near Earth's surface, $g \approx 9.8 \, \text{m/s}^2$.
• As a projectile moves upward, its vertical velocity decreases by $g$ every second until it reaches zero at the peak of its trajectory.
• As it moves downward, its vertical velocity increases by $g$ every second.
• Therefore, the correct answer for the rock thrown upward in section 1 is that its vertical component of velocity C) decreases.

3. Horizontal Motion with Constant Velocity

• The horizontal component of velocity ( $v_x$ ) remains constant throughout the flight, assuming no air resistance. There is no horizontal acceleration.
• Therefore, the correct answer for the rock as it rises in section 2 is that its horizontal component of velocity B) remains unchanged.

4. Acceleration in Projectile Motion

• The acceleration of a projectile is always due to gravity and is constant in both magnitude ($g$) and direction (downwards) at every point in its path (after launch and before landing).
• This means that for section 5, the acceleration at all points (Just after launch, Halfway up, At the peak, Halfway down, Just before landing) is equal in magnitude: A = B = C = D = E.

5. Velocity Vector at Peak

• At the apex (peak) of a projectile's trajectory, its vertical velocity component momentarilly becomes zero ($v_y = 0$).
• The horizontal velocity component ($v_x$) remains, so the velocity vector is B) entirely horizontal at the peak (section 7).

6. Magnitude of the Velocity Vector

• The magnitude of the velocity vector (speed) is not constant because the vertical component of velocity changes due to gravity.
• Speed is minimum at the peak (where $v_y = 0$) and maximum at launch and landing (if at the same height).
• Therefore, for section 6, the statement "As a projectile moves, the magnitude of its velocity vector is constant even though its components change" is B) False. The magnitude changes because $v_y$ changes.

7. Time of Flight, Range, and Initial Angle

• Time of Flight: Depends on the initial vertical component of velocity ($v<em>{oy} = v</em>0 \sin\theta$). A larger $\sin\theta$ for a given initial speed means a longer time in the air.
  • For two cannonballs launched at $30^ riangle$ and $60^ riangle$ (section 3), $60^ riangle$ has a larger initial vertical velocity (\sin 60^ riangle > \sin 30^ riangle), so it stays in the air longer. The $30^ riangle$ cannonball will impact the ground first: A) $30^ riangle$.
• Horizontal Range ($R$): The horizontal distance covered. The formula for range on a flat surface is $R = \frac{v_0^2 \sin(2\theta)}{g}$.
  • For complementary angles (angles that sum to $90^ riangle$), such as $30^ riangle$ and $60^ riangle$, the $\sin(2\theta)$ term will be the same ($\sin(2 \times 30^ riangle) = \sin 60^ riangle$ and $\sin(2 \times 60^ riangle) = \sin 120^ riangle = \sin(180^ riangle - 60^ riangle) = \sin 60^ riangle$).
  • Thus, for section 4, both cannonballs will land at the same point: C) neither; both land at the same point.
• Compared to a Dropped Object:
  • If an object is thrown horizontally from a height (section 10), its initial vertical velocity is $v_{oy} = 0$. It falls in the same time as an object simply dropped from the same height: C) they will hit at the same time.
  • If thrown at an angle above horizontal (section 11), it has an initial upward vertical velocity component, increasing its time of flight compared to a dropped object: A) the ball (hits later).
  • If thrown at an angle below horizontal (section 12), it has an initial downward vertical velocity component, decreasing its time of flight compared to a dropped object: A) the ball (hits first).

8. Kinematic Equations for Projectile Motion

Applying kinematic equations separately for horizontal and vertical motion:

• Vertical Motion:
  • If $v_{oy} = 0$: $\Delta y = \frac{1}{2}gt^2$ or $t = \sqrt{\frac{2 \Delta y}{g}}$
  • General: $\Delta y = v<em>{oy}t + \frac{1}{2}gt^2$, $v</em>y = v<em>{oy} + gt$, $v</em>y^2 = v_{oy}^2 + 2g\Delta y$
• Horizontal Motion:
  • ${\Delta x = v_x t}$
• Airplane Dropping a Load (section 8): The sack maintains the airplane's horizontal velocity. In 4 seconds, ${\Delta x = (40 \, \text{m/s}) \times (4 \, \text{s}) = 160 \, \text{m}}$. So it will be B) 160 m in front of the plane.
• Horizontal Ball Measurement (section 9): To find pitching speed ($v_x$), if you know horizontal displacement ($\Delta x$), you need the time of flight ($t$). This time can often be determined from vertical displacement ($\Delta y$) if the ball is launched horizontally or a known initial vertical velocity. Therefore, A) Either C or E (you need $\Delta y$ or $t$). If you have $\Delta y$, you can calculate $t$.
• Toy Car Off Table (section 14): Find $t$ from $\Delta y = 1.25 \, \text{m}$ (assuming $v<em>{oy} = 0$): $t = \sqrt{\frac{2 \times 1.25\, \text{m}}{9.8 \, \text{m/s}^2}} \approx 0.505 \, \text{s}$. Then find $v</em>x$ from ${\Delta x = v<em>x t}$: $v</em>x = \frac{0.75 \, \text{m}}{0.505 \, \text{s}} \approx 1.48 \, \text{m/s}$.
• Height Doubling Impact (section 15): If $\Delta y$ doubles, time of flight ($t$) increases by a factor of $\sqrt{2}$ (since $t \propto \sqrt{\Delta y}$). As horizontal distance ${\Delta x = v_x t}$, if $t$ increases by $\sqrt{2}$, then $\Delta x$ also increases by $\sqrt{2}$, not doubles. So, the horizontal distance will not double; it will increase by a factor of $\sqrt{2}$.
• Mountain Lion Leap (section 16): Given $\Delta x = 10 \, \text{m}$ and $t = 1.57 \, \text{s}$. Assume symmetrical leap.
  • Horizontal velocity component: $v_x = \frac{\Delta x}{t} = \frac{10 \, \text{m}}{1.57 \, \text{s}} \approx 6.37 \, \text{m/s}$.
  • Time to peak: $t_{peak} = t/2 = 1.57 \, \text{s} / 2 = 0.785 \, \text{s}$.
  • At peak, $v<em>y = 0$. So, $v</em>y = v<em>{oy} - gt</em>{peak} \implies 0 = v_{oy} - (9.8 \, \text{m/s}^2)(0.785 \, \text{s})$.
  • Initial vertical velocity component: $v_{oy} = (9.8 \, \text{m/s}^2)(0.785 \, \text{s}) \approx 7.69 \, \text{m/s}$.
  • Component form of initial velocity: ${\vec{v}_0 = (6.37 \, \text{m/s})\hat{i} + (7.69 \, \text{m/s})\hat{j}}$.
  • Magnitude: |\vec{v}0| = \sqrt{vx^2 + v_{oy}^2} = \sqrt{(6.37)^2 + (7.69)^2} \approx 10.0 \, \text{m/s}}.
  • Direction: $\theta = \arctan\left(\frac{v<em>{oy}}{v</em>x}\right) = \arctan\left(\frac{7.69}{6.37}\right) \approx 50.4^ riangle$ above horizontal.
• Golf Ball on the Moon (section 13): Earth's $g<em>E = 9.8 \, \text{m/s}^2$. Moon's $g</em>M = g<em>E/6 \approx 1.63 \, \text{m/s}^2$. Initial speed $v</em>0 = 25 \, \text{m/s}$, angle $\theta = 30^ riangle$.
  • Range on Earth: $R<em>E = \frac{v</em>0^2 \sin(2\theta)}{g_E} = \frac{(25 \, \text{m/s})^2 \sin(60^ riangle)}{9.8 \, \text{m/s}^2} \approx \frac{625 \times 0.866}{9.8} \approx 55.1 \, \text{m}$.
  • Range on Moon: $R<em>M = \frac{v</em>0^2 \sin(2\theta)}{g_M} = \frac{(25 \, \text{m/s})^2 \sin(60^ riangle)}{1.63 \, \text{m/s}^2} \approx \frac{625 \times 0.866}{1.63} \approx 332.2 \, \text{m}$.
  • Difference: $R<em>M - R</em>E \approx 332.2 \, \text{m} - 55.1 \, \text{m} = 277.1 \, \text{m}$. The ball travels approximately 277.1 m farther on the Moon.

9. Vector Addition and Relative Velocity

• Vectors: Quantities with both magnitude and direction (e.g., displacement, velocity, force, acceleration).
• Components: A vector can be broken down into perpendicular components (e.g., $v<em>x$ and $v</em>y$).
• Algebraic Addition (section 17): For perpendicular vectors, use the Pythagorean theorem for magnitude and tangent function for direction.
  • 15 cm South (y-dir) and 10 cm West (x-dir).
  • Magnitude: $R = \sqrt{(10 \, \text{cm})^2 + (15 \, \text{cm})^2} = \sqrt{100 + 225} = \sqrt{325} \approx 18.03 \, \text{cm}$.
  • Direction: $\theta = \arctan\left(\frac{15 \, \text{cm}}{10 \, \text{cm}}\right) = \arctan(1.5) \approx 56.3^ riangle$ South of West (or 236.3 degrees from East).
• Graphical Addition (Tip-to-Tail) (section 17, 18): Place the tail of the second vector at the tip of the first vector. The resultant vector is drawn from the tail of the first to the tip of the second.
• Relative Velocity (Boat in River) (section 19): When an object moves in a medium that is also moving (like a boat in a river), its resultant velocity is the vector sum of its velocity relative to the medium and the medium's velocity relative to the ground.
  • a. Boat not rowed: Resultant velocity = current velocity = 5 mi/hr East.
  • b. To row straight north: Boat's velocity relative to water (rowing velocity) must have a westward component that cancels the current's eastward component, and a northward component that is the desired speed.
    • Let boat's speed in still water be $v<em>B = 5 \, \text{mi/hr}$, current speed be $v</em>C = 4 \, \text{mi/hr}$ (East).
    • To go straight North, the boat must point somewhat West of North.
    • The westward component of the boat's velocity must be $4 \, \text{mi/hr}$ to cancel the current.
    • Let the angle the boat points from North be $\phi$ (West of North).
    • $v_B \sin\phi = 4 \, \text{mi/hr} \implies (5 \, \text{mi/hr}) \sin\phi = 4 \, \text{mi/hr}$.
    • $\sin\phi = 4/5 = 0.8 \implies \phi \approx 53.1^ riangle$ West of North.
    • The resultant speed (Northward): $v<em>{res} = v</em>B \cos\phi = (5 \, \text{mi/hr}) \cos(53.1^ riangle) \approx (5) \times (0.6) = 3 \, \text{mi/hr}$. So, velocity needed is 3 mi/hr North.
  • c. The angle to point the boat is 53.1$^ riangle$ West of North (as calculated in part b).
    -