Time in air is the duration a projectile spends in free fall, found from the vertical component of motion alone using h = ½gt² for a horizontal launch. Because gravity only acts vertically, launch speed does not change the time; only the height (and gravity) does.
Time in air (also called time of flight) is how long a projectile stays in free fall between launch and landing. For an object launched horizontally, you find it from the vertical motion only. The object starts with zero vertical velocity, accelerates downward at g, and falls a height h, so h = ½gt² gives you t = √(2h/g). Notice what's missing from that equation. The horizontal launch speed appears nowhere.
This is the payoff of the big idea in Topic 1.5. Two-dimensional motion splits into two independent one-dimensional problems. The vertical component has constant acceleration g and controls the clock. The horizontal component has zero acceleration and just coasts at constant velocity for however long the vertical motion allows. A ball fired horizontally off a table and a ball simply dropped from the same height hit the floor at the same instant. The fired ball just lands farther away.
Time in air lives in Topic 1.5 (Vectors and Motion in Two Dimensions) in Unit 1: Kinematics. It directly supports learning objective 1.5.B, which asks you to describe two-dimensional motion by separating it into components, and it leans on 1.5.A, resolving vectors into perpendicular components. The CED's essential knowledge spells out exactly why this works. Projectile motion has zero acceleration in one dimension (horizontal) and constant, nonzero acceleration in the other (vertical). Time in air is the bridge between those two dimensions. You solve the vertical problem to get t, then carry that same t over to the horizontal problem to find range. Almost every projectile calculation on the exam runs through this step, which makes it one of the highest-leverage skills in Unit 1.
Keep studying AP® Physics 1 Unit 1
Range equation (Unit 1)
Range is horizontal velocity times time in air (x = v₀t for a horizontal launch). You can't find how far a projectile travels until you know how long it's airborne, so time in air is step one of nearly every range problem.
Vector components and trigonometry (Unit 1)
For angled launches, you first resolve the initial velocity into components with sin θ and cos θ (LO 1.5.A). Only the vertical component v₀sin θ goes into finding time in air; the horizontal component just waits around to be multiplied by t.
One-dimensional free fall (Unit 1)
Time in air is just a free-fall problem wearing a projectile costume. The same constant-acceleration kinematic equations from earlier in Unit 1 apply to the vertical component, with a = g downward and (for horizontal launches) initial vertical velocity of zero.
No released FRQ uses the phrase 'time in air' as a headline term, but the calculation behind it shows up constantly in projectile questions. Multiple-choice stems love the conceptual trap version, like asking which of two horizontally launched balls lands first when one is faster. The answer is they land together if the height is the same, because time in air depends only on vertical drop. You should be able to (1) compute t = √(2h/g) for a horizontal launch, (2) use that t in the horizontal equation to find range or launch speed, and (3) explain in words why horizontal velocity doesn't affect fall time, citing the independence of perpendicular components. That last skill is exactly what a 'justify your answer' prompt is fishing for.
Time in air answers 'how long?' and range answers 'how far?' Time in air comes from the vertical motion alone and ignores horizontal speed entirely. Range needs both pieces, since it equals horizontal velocity multiplied by time in air. Mixing them up leads to the classic error of thinking a faster horizontal launch makes a projectile stay up longer. It doesn't. It just covers more ground in the same time.
Time in air for a horizontally launched projectile comes entirely from the vertical motion, using h = ½gt², so t = √(2h/g).
Horizontal launch speed has zero effect on time in air; a ball thrown horizontally and a ball dropped from the same height land at the same instant.
Time in air is the link between the two dimensions: solve the vertical problem for t, then plug that t into the horizontal equation to find range.
Projectile motion works this way because acceleration is zero horizontally and constant (g) vertically, the exact setup described in LO 1.5.B.
For angled launches, only the vertical component of initial velocity (v₀sin θ) matters for finding time in air.
Time in air is how long a projectile spends in free fall after launch. For a horizontal launch from height h, it equals √(2h/g), determined only by the drop height and gravitational acceleration.
No. Horizontal velocity does not affect time in air at all, because gravity only acts vertically. A projectile launched at 5 m/s and one launched at 50 m/s from the same height hit the ground at the same time; the faster one just lands farther away.
Time in air is the duration of the flight and depends only on vertical motion. Range is the horizontal distance covered, found by multiplying horizontal velocity by time in air. You need time in air first to calculate range.
Use the vertical equation h = ½gt² with initial vertical velocity zero, then solve for t = √(2h/g). For example, falling from 5 m with g ≈ 10 m/s² gives t = 1 s.
Yes. An angled launch has an initial vertical velocity component (v₀sin θ), so the projectile rises before falling and stays airborne longer than a horizontal launch from the same height. You still find the time using only the vertical component of the motion.
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