Lesson 3 · 16 min
Projectiles
Two independent motions sharing a clock. The range equation is a special case, not a religion.
Split the vector
A launched object feels, to first order, only gravity. Horizontal acceleration is zero; vertical acceleration is −g. The clever move is to stop thinking of “the projectile” and think of two problems that happen to share t.
x(t) = v₀ cosθ · t. y(t) = v₀ sinθ · t − ½gt². Eliminate t and you get a parabola. That shape is why irrigation guns, fire hoses, and ballistic inserts all look the same in silhouette.
Maximum at 45°. Complementary angles (30° and 60°) share a range — different hang times.
Worked example
Drone payload drop
- Given: v₀ = 20 m/s horizontal
- Given: height h = 45 m
- Given: g = 9.81 m/s²
Find: Ground range
- Horizontal launch: θ = 0, so the range equation does not apply.
- Time of fall from y: 0 = 45 − ½gt² → t = √(2h/g) = √(90/9.81) ≈ 3.03 s
- x = v₀ t = 20 × 3.03
61 m (release 61 m before the target, in still air)
Check
Why does a 30° launch and a 60° launch land at the same range on level ground (no drag)?
Bench
Projectile
R = v² sin(2θ) / g
- Range
- 48.6 m
- Hang
- 2.88 s
- H max
- 10.2 m
- Target
- 13.6 m off