Motion in two directions
| English | Chinese | Pinyin |
|---|---|---|
| vertical | 竖直 | shù zhí |
| horizontal | 水平 | shuǐ píng |
| parabola | 抛物线 | pāo wù xiàn |
| range | 射程 | shè chéng |
| symmetric | 对称 | duì chèn |
| projectile | 抛体 | pāo tǐ |
Drop one, throw one
- Drop a ball and throw another one sideways at the same instant.
- They hit the ground at the same time.
- The sideways motion does not change how fast it falls.
Two motions, side by side
- Horizontal 水平 and vertical 竖直 motion are independent.
- Treat each direction as its own SUVAT problem, linked only by the time $t$.
A high-speed train: its motion can be shown on a distance-time graph
Launch a projectile
Fire the ball, then change the angle and speed. The horizontal motion is steady while gravity pulls it down — together they trace a parabola. Find the angle for the longest range, and try the Moon.
A horizontal throw
- Horizontal: constant velocity $u_{\text{H}}$, so $x = u_{\text{H}}t$.
- Vertical: free fall from rest, so $y = \tfrac{1}{2}gt^{2}$.
- The time to land depends only on the height, not on $u_{\text{H}}$.
Displacement–time graph of a car on a test track
A ball thrown horizontally and a ball dropped from the same height hit the ground at the same time (no air resistance).
Yes — the vertical motion is the same free fall for both, and the horizontal speed does not change the fall time.
Launched at an angle
- Split the launch velocity: horizontal $u\cos\theta$ (stays constant), vertical $u\sin\theta$.
- The vertical part slows, stops at the top, then grows downward — tracing a parabola 抛物线.
A projectile is launched at $20\ \dfrac{\text{m}}{\text{s}}$, $30^{\circ}$ above the horizontal. What is the vertical part of the launch velocity?
Vertical part $= u\sin\theta = 20 \times \sin 30^{\circ} = 20 \times 0.5 = 10\ \dfrac{\text{m}}{\text{s}}$.
At the top and the range 射程
- At the highest point $v_{\text{V}} = 0$, but $v_{\text{H}} = u\cos\theta$ is unchanged.
- Time to the top $= \dfrac{u\sin\theta}{g}$; the flight is symmetric 对称, so total time is twice that.
Projectile 抛体 launched at angle $\theta$ — horizontal and vertical motions are independent
Select all the statements that are true for a projectile (ignore air).
At the top only the vertical velocity is zero — the horizontal velocity $u\cos\theta$ carries on, so the ball keeps moving. Mass does not change free fall.
That projectile has a vertical launch velocity of $10\ \dfrac{\text{m}}{\text{s}}$. How long does it take to reach the top? (Use $g = 9.81\ \dfrac{\text{m}}{\text{s}^2}$.)
Time to the top $= \dfrac{u\sin\theta}{g} = \dfrac{10}{9.81} \approx 1.0\ \text{s}$.
A bouncing ball
- Between bounces the velocity–time graph is a straight sloping line (constant $g$).
- Each bounce is a sudden jump: the velocity flips direction (and shrinks if energy is lost).
On the velocity–time graph of a bouncing ball, each bounce appears as:
At a bounce the velocity flips from downward to upward almost instantly — a sudden vertical jump on the graph (smaller if energy is lost).
Two objects meeting
- Write a displacement equation for each, using the same start time and positive direction.
- They are level again when their displacements are equal — set $s_1 = s_2$ and solve.
Two objects moving along the same line are level again when their ____ are equal.
Set the two displacement equations equal ($s_1 = s_2$) and solve for the time.
You've got it
- horizontal and vertical motion are independent — share only the time $t$
- a horizontal throw lands in a time set only by its height
- resolve a launch: $u\cos\theta$ stays constant, $u\sin\theta$ behaves like a vertical throw