Horizontal Range and Complementary Angles
For ideal projection on level ground, R = u squared sin 2theta divided by g. The maximum is u squared over g at 45 degrees, and theta and 90 degrees minus theta have equal ranges.
Why this shows up in the exam
Choosing launch angles for a target range · Comparing complementary trajectories · Inferring launch speed from maximum reach
Learn the idea
For equal launch and landing levels, range depends on sin twice the launch angle. A shallow shot travels quickly for a short time, while a steep shot travels slowly sideways for longer. Complementary angles balance those effects and give the same range.
🧠 Memory hook: Range sees twice the angle; complementary launches share a range.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- R = u² sin(2theta)/g — level-ground horizontal range
- R_max = u²/g at theta = 45 degrees — maximum range for fixed speed
- R = u cos(theta) T — range from horizontal speed and flight time
How to approach it
- 1Confirm launch and landing levels match
- 2Apply the range formula or combine T with horizontal speed
- 3Check for complementary-angle alternatives
Common slip-ups that cost marks
- •Using the level-ground formula for unequal heights
- •Claiming 45 degrees is always optimal
- •Forgetting the second complementary solution
🌟 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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