Horizontal Projection and Dropped Bodies
For horizontal launch speed u_x from height h over level ground, t = sqrt(2h/g), horizontal displacement is u_x t, and impact velocity combines u_x with gt downward.
Why this shows up in the exam
Air-drop targeting · Objects rolling off tables or stairs · Comparing drops from moving vehicles
Learn the idea
A horizontally released body keeps the carrier's horizontal velocity while gravity builds vertical velocity. A package released from a moving aircraft does not lose its forward speed. Its fall time comes from vertical drop, and that same time sets the horizontal lead distance.
🧠 Memory hook: Drop decides the time; carried horizontal speed decides the lead.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- t_fall = sqrt(2h/g) — fall time from rest vertically
- x = u_x sqrt(2h/g) — horizontal displacement during the fall
- v_impact = sqrt(u_x² + 2gh) — impact speed without air resistance
How to approach it
- 1Solve vertical fall for time
- 2Carry the release horizontal velocity unchanged
- 3Combine impact components with correct direction
Common slip-ups that cost marks
- •Making the released object stop horizontally
- •Using horizontal distance to find fall time first
- •Including mass in ideal fall time
🌟 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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