Projectile Components and Instantaneous Velocity
For launch speed u at angle theta above horizontal with no air resistance, v_x = u cos theta and v_y = u sin theta - gt; position follows by integrating these components.
Why this shows up in the exam
Finding velocity at a specified time · Determining flight direction or kinetic energy · Comparing ascent and descent states
Learn the idea
Ideal projectile motion is uniform horizontally and uniformly accelerated vertically. After launch, gravity changes only the vertical velocity while the horizontal component stays fixed. Every standard result comes from keeping those two component stories synchronized.
🧠 Memory hook: Horizontal keeps its launch value; gravity edits only vertical velocity.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- v_x = u cos(theta) — constant horizontal velocity
- v_y = u sin(theta) - gt — vertical velocity at time t
- v = sqrt(v_x² + v_y²) — instantaneous speed
How to approach it
- 1Resolve the launch velocity
- 2Update only the vertical component with time
- 3Recombine components for speed, direction, or momentum
Common slip-ups that cost marks
- •Reducing horizontal speed because of gravity
- •Using launch angle as the later velocity angle
- •Dropping the sign of vertical velocity during descent
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A particle has initial speed 2 m/s and constant acceleration 3 m/s^2 for 4 s. What distance does it cover?
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