Projectile Trajectory Equation and Geometric Constraints
For launch from the origin under uniform downward gravity, y = x tan theta - gx squared divided by 2u squared cos squared theta. A shifted origin adds the initial coordinates.
Why this shows up in the exam
Checking whether a projectile clears a wall · Finding launch speed and angle from a path equation · Testing reachability of a target point
Learn the idea
Eliminating time turns the projectile's component equations into a parabola. The same flight can be described by where the projectile is at each time or by the shape y(x). The trajectory form is especially useful for walls, target points, and extracting launch parameters.
🧠 Memory hook: Eliminate time and the two component motions become one parabola.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- y = x tan(theta) - g x²/[2u² cos²(theta)] — ideal projectile trajectory from the origin
- tan(theta) = initial slope dy/dx at x=0 — launch angle from trajectory slope
- H = alpha²/(4 beta) for y = alpha x - beta x² — vertex height of a parabolic trajectory
How to approach it
- 1Set a convenient origin
- 2Write x(t) and y(t), then eliminate t
- 3Apply the stated point, wall, or vertex constraint
Common slip-ups that cost marks
- •Using the formula with a shifted origin unchanged
- •Reading beta as gravity alone
- •Ignoring that a target point may allow two launch angles
🌟 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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