Maximum Height and the Highest Point
For ideal launch and height measured above the launch level, H = u squared sin squared theta divided by 2g, reached at t = u sin theta/g.
Why this shows up in the exam
Finding clearance above obstacles · Comparing launch angles by peak height · Relating top speed to launch energy
Learn the idea
At the highest point vertical velocity is zero, but horizontal velocity generally remains. Gravity gradually removes the upward velocity component. At the top the projectile is still moving sideways, so neither its speed nor kinetic energy is usually zero.
🧠 Memory hook: At the top, vertical stops for an instant; horizontal does not.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- v_y = 0 at the highest point — vertical turning-point condition
- H = u² sin²(theta)/(2g) — maximum height above launch level
- v_top = |u cos(theta)| — speed at the top without air resistance
How to approach it
- 1Set vertical velocity to zero
- 2Use vertical kinematics for height or ascent time
- 3Retain the horizontal component for speed and momentum
Common slip-ups that cost marks
- •Setting total velocity to zero at the top
- •Measuring height from the wrong reference level
- •Using sin theta rather than sin squared theta
🌟 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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