Projectile motion and variable acceleration
แนะนำใหม่| English | ความหมาย |
|---|---|
| projectile/prəˈdʒektaɪl/ | วัตถุที่เคลื่อนที่ตามเส้นโค้งในอากาศ |
Why does horizontal motion share vertical time?
- A ball is launched at an angle. Its horizontal and vertical motion share time but follow different equations.
- This lesson studies projectile 抛体: A particle moving under gravity after launch in a model that neglects air resistance.
The method
- Resolve initial velocity into horizontal u cosθ and vertical u sinθ.
- With no air resistance, horizontal acceleration is zero and vertical acceleration is -g.
- Use the same time in both components.
At the highest point of a projectile’s flight, which is true?
ความยาก 1/5 · 100 เหรียญ + โบนัสเล็กน้อยในการทำครั้งแรกวันนี้
Gravity acts only vertically, so the horizontal component stays constant throughout. At the peak v_y=0 but v_x keeps its launch value, and the acceleration remains g downward.
A worked case
For components u_x=12,u_y=16 and g=9.8, at t=2 s the horizontal displacement is x=12×2=24 m and vertical displacement is y=16×2-9.8×2²/2=12.4 m. At the highest point, v_y=0 but v_x remains 12.
Projectile motion and variable acceleration
Resolve initial velocity into horizontal u cosθ and vertical u sinθ
Separate the two components and interpret the sign of the vertical velocity.
For u_x=12,u_y=16,g=9.8, find x at t=2.
ความยาก 2/5 · 140 เหรียญ + โบนัสเล็กน้อยในการทำครั้งแรกวันนี้
Horizontal motion is uniform: x=12×2=24 m.
Where it goes wrong
- Zero vertical velocity at the highest point does not mean zero total speed.
- Launch and landing heights need not be equal.
- Do not use a range formula that assumes equal heights without checking them.
Not true: The total speed of a projectile is zero at its highest point. At the top the vertical velocity is zero, but the horizontal velocity remains, so the speed is not zero.
For that launch, find y at t=2.
ความยาก 2/5 · 140 เหรียญ + โบนัสเล็กน้อยในการทำครั้งแรกวันนี้
y=16×2-9.8×2²/2=12.4 m.
At the top the vertical velocity is zero, but the horizontal velocity remains, so the speed is not zero.
ความยาก 1/5 · 100 เหรียญ + โบนัสเล็กน้อยในการทำครั้งแรกวันนี้
True. At the top the vertical velocity is zero, but the horizontal velocity remains, so the speed is not zero.
Applying it
- For variable acceleration, integrate a(t) to get v(t) and use the initial velocity to find the constant, then integrate for displacement.
- Confirm units and the physical time interval.
Find v_y at t=2.
ความยาก 2/5 · 140 เหรียญ + โบนัสเล็กน้อยในการทำครั้งแรกวันนี้
v_y=16-9.8×2=-3.6 m/s, so motion is downward.
Match each projectile component to its behaviour without air resistance.
ความยาก 2/5 · 140 เหรียญ + โบนัสเล็กน้อยในการทำครั้งแรกวันนี้
The two components are independent motions sharing the same time variable; resolve first, then apply constant-acceleration formulae separately.
In your course
- Pearson Edexcel International A Level Further Mathematics, unit M2.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Projectile: A particle moving under gravity after launch in a model that neglects air resistance.
- Resolve initial velocity into horizontal u cosθ and vertical u sinθ.
- Watch for: zero vertical velocity at the highest point does not mean zero total speed.
Write from a horizontal distance of 10 m
- Horizontal speed is 5 m/s and the flight lasts 2 s. Ignore air resistance.
- Find the horizontal distance.
A projectile has constant horizontal speed 5 m/s and is in the air for 2 s. What horizontal distance does it travel?
5 times 2 is 10.
What is the horizontal distance?
ความยาก 1/5 · 100 เหรียญ + โบนัสเล็กน้อยในการทำครั้งแรกวันนี้
Constant horizontal speed times time is 10 m. Other options — 24 m (the lesson value); 5 m (the speed used as a distance).