Motion of a projectile
مُقدَّم| English | المعنى |
|---|---|
| horizontal/ˌhɒrɪˈzɒntl/ | أفقي |
| vertical/ˈvɜːtɪkl/ | عمودي |
| gravity/ˈɡrævɪti/ | الجاذبية |
| projectile/prəˈdʒektaɪl/ | قذيفة |
| parabola/pəˈræbələ/ | قطع مكافئ |
| range/reɪndʒ/ | المدى الأفقي |
| greatest height/ˈɡreɪtɪst haɪt/ | الارتفاع الأقصى |
The cannonball problem
- Fire a ball at an angle and it traces a smooth arch, then lands. Where, and how high?
- Galileo's insight: treat the horizontal 水平 and vertical 竖直 motions completely separately — they only share the clock.
Two independent motions
- Launched at speed $u$ and angle $\alpha$:
- Horizontal: constant velocity $u\cos\alpha$ (no force sideways).
- Vertical: gravity 重力 only, acceleration $g$ downward.
Water jets follow parabolic paths, the classic projectile 抛射体 motion under gravity
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.
During the flight, the horizontal velocity of a projectile (ignoring air resistance):
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There is no horizontal force, so u cos α is constant throughout.
The initial horizontal velocity of a projectile launched at speed u and angle α is u ______ α.
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Horizontal component = u cos α; vertical = u sin α.
The parabola 抛物线
- Eliminating $t$ leaves $y$ as a quadratic in $x$ — so the path is a parabola.
Horizontal speed stays constant; gravity bends the vertical motion into a parabola.
The path of a projectile (no air resistance) is a:
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Eliminating t gives y as a quadratic in x — a parabola.
Range 射程 and greatest height 最大高度
- At the top, the vertical velocity is momentarily zero. The key results:
Worked example. $u = 20\ \tfrac{\text{m}}{\text{s}}$, $\alpha = 30^\circ$, $g = 10$. Greatest height $= \dfrac{400 \times 0.25}{20} = 5\ \text{m}$.
A ball is thrown at u = 20 m/s at 30°. Using height = u²sin²α/(2g) with g = 10, find the greatest height (m).
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400 × sin²30° / 20 = 400 × 0.25 / 20 = 5 m.
At what launch angle (degrees) is the range greatest?
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Range ∝ sin 2α, which is largest when 2α = 90°, i.e. α = 45°.
At the highest point of a projectile's flight, its velocity is zero.
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False. Only the vertical component is zero at the top; the horizontal component is unchanged, so the speed is not zero.
Common trap
Range is maximised at $45^\circ$, not $90^\circ$. $\sin 2\alpha$ peaks when $2\alpha = 90^\circ$. Firing straight up gives zero range — all the speed goes into height.
- Eliminating time gives the Cartesian equation of the path; all trajectories lie under a bounding parabola.
You've got it
- resolve into horizontal ($u\cos\alpha$, constant) and vertical ($u\sin\alpha$, gravity) parts
- the path is a parabola
- range $= \dfrac{u^2\sin 2\alpha}{g}$ (max at $45^\circ$); greatest height $= \dfrac{u^2\sin^2\alpha}{2g}$
Write from a greatest height of 20 m
- A ball is thrown at 40 m/s at 30 degrees. Take g = 10.
- Find the greatest height. The range is greatest at 45 degrees, not 90.
A ball is thrown at 40 metres per second at 30 degrees, with g = 10. What is the greatest height, and why is the range not greatest at 90 degrees?
The greatest height is u squared sin squared alpha divided by 2g. The range formula peaks when the double angle is 90 degrees.
What is the greatest height?
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1600 divided by 4 is 400, and 400 divided by 20 is 20 metres. Other options — 5 metres (the lesson height for 20 metres per second); 45 degrees (the angle for the greatest range).