Notes
Vectors
- Scalars have magnitude only; vectors have magnitude and direction.
- Perpendicular vectors: $\displaystyle R = \sqrt{A^2 + B^2}$ and $\displaystyle \theta = \tan^{-1}\left(\frac{B}{A}\right)$.
- Components: $V_x = V\cos\theta$, $V_y = V\sin\theta$. Add components to add vectors.
Projectile motion
- Horizontal and vertical motions are independent. Horizontal: constant $v_x$. Vertical: free fall with $g = 9.8 \text{ m/s}^2$.
- Horizontal launch: time from $\displaystyle y = \frac{1}{2}gt^2$, then $x = v_x t$.
- Angled launch on level ground: total time $\displaystyle = \frac{2v_y}{g}$, range $= v_x t$. At the top, $v_y = 0$. Maximum range at $45^\circ$.
Relative velocity
- Add velocities of the object relative to the medium and the medium relative to the ground.
- Crossing time depends only on the velocity component across the river.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$\displaystyle \sqrt{81 + 144} = \sqrt{225} = 15 \text{ m}$.
$50\sin 60^\circ = 50(0.866) \approx 43.3 \text{ N}$.
$\displaystyle t = \sqrt{\frac{2(4.9)}{9.8}} = 1 \text{ s}$, so $x = 3 \times 1 = 3 \text{ m}$.
No horizontal force acts, so $v_x$ never changes.
Total time $\displaystyle = \frac{2(9.8)}{9.8} = 2 \text{ s}$.
Opposite directions subtract: $20 - 1.5 = 18.5 \text{ m/s}$ forward.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A soccer player kicks a ball from level ground at $25 \text{ m/s}$ at an angle of $37^\circ$ above the horizontal. Use $\sin 37^\circ = 0.60$, $\cos 37^\circ = 0.80$, $g = 9.8 \text{ m/s}^2$, and ignore air resistance.
(a) Find the horizontal and vertical components of the initial velocity.
Show the answer and scoring guide
$v_x = 25(0.80) = 20 \text{ m/s}$; $v_y = 25(0.60) = 15 \text{ m/s}$.
Check off the points you earned:
(b) How long is the ball in the air?
Show the answer and scoring guide
Time to the top $\displaystyle = \frac{15}{9.8} \approx 1.53 \text{ s}$. Total time $= 2 \times 1.53 \approx 3.06 \text{ s}$.
Check off the points you earned:
(c) How far away does the ball land, and what is its maximum height?
Show the answer and scoring guide
Range $= v_x t = 20 \times 3.06 \approx 61 \text{ m}$. Maximum height from $0 = v_y^2 - 2gh$: $\displaystyle h = \frac{15^2}{2(9.8)} \approx 11.5 \text{ m}$.
Check off the points you earned: