Projectiles Launched at an Angle
- Resolve a launch velocity into independent horizontal and vertical components
- Use the symmetry of the parabolic path to shortcut time-of-flight and speed questions
- Derive the range equation and explain why 45° maximizes range on level ground
The only new step is the first one
An angled launch is the horizontal-launch problem with one extra line of setup. Split the launch velocity into components — v₀ₓ = v₀ cos θ and v₀ᵧ = v₀ sin θ — and then the two axes proceed exactly as before. Horizontally, no acceleration and constant velocity. Vertically, a constant-acceleration problem, now with a nonzero starting velocity. Time remains the only quantity shared between them.
At the top, vᵧ = 0 — but the object is still moving
At the apex the vertical velocity is momentarily zero, which is what makes the apex easy to find: set vᵧ = 0 in v = v₀ᵧ − gt and solve for time. But the object is not at rest there. Its horizontal velocity is unchanged, so its speed at the top equals v₀ₓ = v₀ cos θ. And its acceleration at the top is still g downward — the single most-missed idea in this unit. Zero velocity does not mean zero acceleration; if acceleration were zero at the apex the object would hang there forever.
Symmetry is a shortcut, not a coincidence
For a projectile that lands at the same height it was launched from, the path is symmetric about the apex. Three consequences save real time on the exam: the time up equals the time down, so total flight time is twice the rise time; the landing speed equals the launch speed; and the landing angle below horizontal equals the launch angle above it. All three follow from the vertical motion being the same constant-acceleration problem run forward and then backward.
A ball is launched at 20 m/s at 30° above the horizontal from level ground. Using g = 10 m/s², find the time of flight, the maximum height and the range.
- 1.Components: v₀ₓ = 20 cos 30° = 17.3 m/s, v₀ᵧ = 20 sin 30° = 10 m/s.
- 2.Time to apex from vᵧ = 0 = v₀ᵧ − gt: t = 10 ÷ 10 = 1.0 s. Level ground, so total flight = 2.0 s.
- 3.Maximum height: Δy = v₀ᵧt − ½gt² = (10)(1.0) − ½(10)(1.0)² = 10 − 5 = 5.0 m.
- 4.Range: Δx = v₀ₓ × total time = (17.3)(2.0) = 34.6 m.
Complementary angles give the same range
Because sin(2θ) appears in the range equation, launching at 30° and at 60° with the same speed produces the same range on level ground — sin 60° and sin 120° are equal. The trajectories are not the same, though: the 60° shot goes much higher and stays airborne much longer. It is a favorite multiple-choice item precisely because "same range" and "same path" feel like they should go together.
The range equation is a derived special case, not a fundamental law. It fails the moment the projectile lands at a different height than it was launched from — off a cliff, into a basket, onto a roof. In those problems, go back to the component equations.
A projectile is launched at 25 m/s at 53° above the horizontal. What is its speed at the highest point of its flight? (cos 53° ≈ 0.60)
A ball launched from level ground takes 3.0 s to return to the ground. How long after launch does it reach its maximum height? (Ignore air resistance.)
Two balls are launched from level ground with equal speeds, one at 25° and one at 65°. Compared with each other, they have:
Write "up is positive" at the top of any projectile free-response and use it consistently. Rubrics award the sign convention, and a single flipped sign in the vertical equation usually propagates into every later part of the question.
Answer the 3 checkpoints as you read.
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