Projectile motion and variable acceleration
| English | Français |
|---|---|
| projectile/prəˈdʒektaɪl/ | projectile |
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.
Choose the mathematical structure
- 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.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines projectile?
A particle moving under gravity after launch in a model that neglects air resistance.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
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.
Horizontal motion is uniform: x=12×2=24 m.
Test a tempting shortcut
- 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.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
The total speed of a projectile is zero at its highest point. This claim is false. Explain which definition or assumption it violates.
For that launch, find y at t=2.
y=16×2-9.8×2²/2=12.4 m.
The total speed of a projectile is zero at its highest point.
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.
Interpret a new situation
- 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.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find v_y at t=2.
v_y=16-9.8×2=-3.6 m/s, so motion is downward.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- edexcel IAL further mathematics; official unit M2. Other-unit enrichment is identified in the scope review; it is not an extra cash-in requirement.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
A particle moving under gravity after launch in a model that neglects air resistance. Choose the relationship, show the method, check its assumptions and interpret the result.