Vertical Projection and Maximum Height
For upward launch speed u with negligible drag, time to the top is u/g and rise is u²/(2g). At equal heights on ascent and descent, speed magnitudes are equal while velocity signs are opposite.
Why this shows up in the exam
Time at a given height · Cliff launches · Velocity-time graphs of vertical motion
Learn the idea
At the highest point of an upward projection, velocity is momentarily zero but acceleration remains downward. Gravity steadily removes upward velocity until the body pauses at the top, then builds downward velocity. The pause does not switch gravity off.
🧠 Memory hook: At the top v is zero, not g.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- t_top = u/g — time from launch to maximum height
- H = u²/(2g) — maximum rise above launch point
- v_top = 0, a_top = -g — top-point velocity and acceleration with upward positive
How to approach it
- 1Use v=0 only at maximum height
- 2Keep launch and landing levels explicit
- 3Solve the height equation for all valid times
Common slip-ups that cost marks
- •Setting acceleration to zero at the top
- •Assuming ascent and descent times are equal when landing height differs
- •Using speed sign instead of velocity sign
🌟 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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