To solve a projectile motion problem, split the motion into two independent parts: horizontally the velocity stays constant (x = vₓt), and vertically the object accelerates downward at g = 9.8 m/s² (y = y₀ + v_y0·t − ½gt²). Time is the only quantity the two parts share, so find the time in the air from the vertical motion, then use it for the horizontal distance. The shortcut R = v₀² sin 2θ / g only works when the projectile lands at the same height it was launched from.
Below are the three launch types you will meet: horizontal off a cliff, angled on level ground, and the case most textbooks skip, angled from a cliff, where the time comes from the quadratic formula. One sign convention and one x/y table cover them all.
The Two Rules Behind Every Projectile Problem
vₓ = v₀ cos θ (constant), x = vₓtVertical:
v_y0 = v₀ sin θ, v_y = v_y0 − gt, y = y₀ + v_y0·t − ½gt²Valid when air resistance is negligible and θ is measured from the horizontal.
A projectile is anything launched and then left to move under gravity alone: a kicked ball, a stone off a cliff. Once it is in the air, the only force on it (ignoring air) is its weight, which points straight down. Nothing pushes it sideways, so its horizontal velocity never changes. Vertically it is in free fall, exactly like a ball thrown straight up, so the kinematic equations apply with a = −g.
This is why the two directions are independent. A ball pushed off a table and a ball dropped from the same height at the same moment hit the floor together, because sideways speed has no effect on how fast the height falls. The trajectory is a parabola because x grows linearly in time while y changes quadratically.
We take g = 9.8 m/s² throughout. The defined standard value is 9.80665 m/s² (NIST), and 9.8 is the usual classroom figure; if your course uses 9.81 or 10, last digits will differ slightly. The data in the examples has three significant figures, so the answers do too.
A 5-Step Method for Any Launch
- Sketch, and put the origin at the launch point
Up and forward are positive. A landing point below the launch has a negative
y, a landing point above it has a positivey. - Split the launch velocity
vₓ = v₀ cos θandv_y0 = v₀ sin θ, using SOH CAH TOA on the velocity arrow. A horizontal launch has θ = 0, sov_y0 = 0. A launch below the horizontal has a negativev_y0. - Fill in an x/y table
Write what you know in two columns: horizontal (
vₓ,a = 0) and vertical (v_y0,a = −9.8 m/s², displacement). Puttacross both columns, because it is shared. - Get t from the column that has enough information
Usually that is the vertical column, since you know the height change. Sometimes the horizontal column gives it instead (you know how far away a wall is).
- Use t in the other column
Then answer what is asked. A landing velocity is rebuilt from
vₓandv_ywith the Pythagorean theorem.
Horizontal Launch Off a Cliff or Table
When something leaves a surface moving horizontally, all of its starting velocity is in vₓ and v_y0 = 0. The fall height alone fixes the time: from −h = −½gt², t = √(2h/g).
A stone is kicked horizontally at 8.00 m/s off a cliff 19.6 m high. How long is it in the air, how far from the base does it land, and how fast is it moving when it lands?
Horizontal: vₓ = 8.00 m/s. Vertical: v_y0 = 0, y = −19.6 m, a = −9.8 m/s².
Time from the vertical column: t = √(2 × 19.6 m ÷ 9.8 m/s²) = √(4.00 s²) = 2.00 s. The metres cancel, leaving s² under the root.
Distance from the horizontal column: x = (8.00 m/s)(2.00 s) = 16.0 m.
Landing velocity: v_y = 0 − (9.8 m/s²)(2.00 s) = −19.6 m/s, and vₓ is still 8.00 m/s. Speed = √(8.00² + 19.6²) = √448.16 = 21.2 m/s, at tan⁻¹(19.6 ÷ 8.00) = 67.8° below the horizontal.
t = 2.00 s, x = 16.0 m, 21.2 m/s at 67.8° below horizontalKick the stone twice as hard and it lands twice as far away, but after exactly the same 2.00 s. The horizontal speed changes where it lands, never when.
Angled Launch on Level Ground
When the projectile lands at its launch height, the flight is symmetric: it takes as long to rise to the top as to fall back. At the top, v_y = 0, but the projectile is still moving at vₓ. Setting y = 0 in the vertical equation gives t(v_y0 − ½gt) = 0, so the time of flight is T = 2v_y0 / g.
A ball is kicked at 20.0 m/s at 30.0° above the horizontal on flat ground. Find the time of flight, the maximum height and the range.
Components: vₓ = 20.0 cos 30.0° = 17.32 m/s, v_y0 = 20.0 sin 30.0° = 10.0 m/s.
Time of flight: T = 2(10.0 m/s) ÷ 9.8 m/s² = 2.04 s.
Maximum height, where v_y = 0: from v_y² = v_y0² − 2gH, H = (10.0 m/s)² ÷ (2 × 9.8 m/s²) = 100 m²/s² ÷ 19.6 m/s² = 5.10 m.
Range: x = (17.32 m/s)(2.041 s) = 35.3 m, using the unrounded time.
T = 2.04 s, H = 5.10 m, R = 35.3 mThe level-ground shortcuts, and why 45° goes farthest
Put T = 2v₀ sin θ / g into x = (v₀ cos θ)T and use 2 sin θ cos θ = sin 2θ (one of the double-angle identities). You get the range equation that OpenStax College Physics 2e, section 3.4 gives for a projectile on level ground with negligible air resistance:
| Quantity (level ground only) | Formula | Example 2 check |
|---|---|---|
| Time of flight | T = 2v₀ sin θ / g | 2.04 s |
| Maximum height | H = (v₀ sin θ)² / 2g | 5.10 m |
| Range | R = v₀² sin 2θ / g | 400 × sin 60° ÷ 9.8 = 35.3 m |
Because sin 2θ is largest (equal to 1) when 2θ = 90°, the longest range for a given speed comes at θ = 45°. And since sin(180° − 2θ) = sin 2θ, two launch angles that add to 90° (like 30° and 60°) give the same range: the steeper one goes higher and stays up longer, but moves sideways more slowly. Both facts are stated in the same OpenStax section, which also notes that with air resistance the best angle drops to about 38°.
Using R = v₀² sin 2θ / g for a launch from a cliff, a roof or a person's hand above the ground. The formula was built by setting the landing height equal to the launch height, so anywhere else it gives the wrong distance. In Example 3 below it would give 35.3 m when the real answer is 64.0 m. If the heights differ, go back to the two rules and solve for t.
Launch From a Height at an Angle
This is the typical AP and exam case. The vertical equation now has all three terms, y = v_y0·t − ½gt² with a negative y, so it is a quadratic in t. Rearrange to ½gt² − v_y0·t + y = 0 and use the quadratic formula. One root is positive and one is negative; the negative one is the time the parabola would have passed ground level before the launch, so it has no physical meaning here.
A ball is thrown at 20.0 m/s at 30.0° above the horizontal from the top of a cliff 30.0 m high. How far from the base of the cliff does it land, and how fast is it going?
Same components as Example 2: vₓ = 17.32 m/s, v_y0 = 10.0 m/s. The landing point is below the launch, so y = −30.0 m.
Vertical: −30.0 = 10.0t − 4.9t², so 4.9t² − 10.0t − 30.0 = 0 (units: m, with t in s).
t = [10.0 ± √(10.0² + 4 × 4.9 × 30.0)] ÷ (2 × 4.9) = (10.0 ± √688) ÷ 9.8 = (10.0 ± 26.23) ÷ 9.8. The roots are 3.697 s and −1.656 s; keep the positive one.
Horizontal: x = (17.32 m/s)(3.697 s) = 64.0 m.
Landing velocity: v_y = 10.0 − 9.8 × 3.697 = −26.23 m/s, so the speed is √(17.32² + 26.23²) = √988 = 31.4 m/s.
t = 3.70 s, x = 64.0 m, landing speed 31.4 m/sThe kinematics calculator handles the vertical step for you if you want to check it: enter u = 10, a = −9.8 and s = −30, and it lists both roots for the time, so you can see the rejected one too.
Landing Higher Than the Launch: Which Root to Keep
When the target is above the launch point, both roots of the quadratic are positive. The projectile passes that height twice: once on the way up and once on the way down. The physics decides which one you want.
A ball is thrown at 20.0 m/s at 60.0° above the horizontal and lands on a flat roof 10.0 m above the launch point. How long is it in the air and how far does it travel horizontally?
Components: vₓ = 20.0 cos 60.0° = 10.0 m/s, v_y0 = 20.0 sin 60.0° = 17.32 m/s.
Vertical: 10.0 = 17.32t − 4.9t², so 4.9t² − 17.32t + 10.0 = 0.
t = (17.32 ± √(300 − 196)) ÷ 9.8 = (17.32 ± 10.20) ÷ 9.8, giving 0.727 s and 2.808 s.
At 0.727 s the ball is still rising (v_y = 17.32 − 9.8 × 0.727 = +10.2 m/s), so it is passing the roof's edge height on the way up and can't land yet. It lands on the way down, at the later root, when v_y = −10.2 m/s.
x = (10.0 m/s)(2.808 s) = 28.1 m.
t = 2.81 s, x = 28.1 mA negative root is before launch, so throw it away. With two positive roots, the smaller is the way up and the larger, where any landing happens, is the way down. Clearing a wall or hitting a target in mid-air can mean either, so check the sign of v_y at each time. If the number under the square root is negative, the projectile never reaches that height at all.
More Mistakes That Cost Marks
- Swapping sine and cosine. With θ measured from the horizontal, the horizontal part uses cos and the vertical part uses sin. If the problem gives the angle from the vertical, they swap, so redraw the triangle rather than memorising.
- Saying the speed is zero at the top. Only
v_yis zero at the peak. The projectile still moves atvₓ, which is also its slowest speed in the flight. - Putting g in the horizontal equation. Gravity has no horizontal part, so
aₓ = 0. If your horizontal velocity changes, something has gone wrong. - Using sin θ where the range formula needs sin 2θ. For Example 2 that slip gives
400 × sin 30° ÷ 9.8 = 20.4 minstead of 35.3 m. - Forgetting the sign of y. Landing 30.0 m below the launch means
y = −30.0 m. Entering +30.0 gives the time to reach a height the ball never gets to, and a negative number under the square root.
How to Check Your Answer
Units first. Every time should come out in seconds, every distance in metres. In t = √(2h/g), metres divided by m/s² leaves s² under the root, which is a quick way to catch a formula you've misremembered.
Then a speed check that ignores the angle. For any projectile without air resistance, the landing speed depends only on the launch speed and the drop: v² = v₀² + 2gh, where h is how far below the launch it lands. This comes from adding vₓ² (unchanged) to v_y² = v_y0² + 2gh.
In Example 3, v = √(20.0² + 2 × 9.8 × 30.0) = √988 = 31.4 m/s, matching the answer from the components. In Example 1, √(8.00² + 2 × 9.8 × 19.6) = 21.2 m/s. If your component answer disagrees with this one-line version, one of them has an arithmetic slip.
Finally, a size estimate. A launch from a height should go farther than the same launch on level ground (64.0 m vs 35.3 m). If it doesn't, look for a sign error in y.
Once you've worked a problem by hand, the physics solver is a way to compare your steps: in Solver AI on iPhone or iPad you can snap a photo of a projectile question and get a step-by-step solution, then ask the AI tutor why a step works if yours differs.
Practice Problems
Use g = 9.8 m/s² and no air resistance.
- A ball rolls off a table
0.800 mhigh at2.50 m/s. How long until it hits the floor, and how far from the table edge does it land? - A ball is launched at
30.0 m/sat45.0°on level ground. Find its time of flight, maximum height and range. - Show that launches at
15.0°and75.0°with a speed of30.0 m/son level ground have the same range, and find it. - A ball is kicked at
18.0 m/sat40.0°toward a fence25.0 maway and3.00 mtall, from ground level. Does it clear the fence? (Hint: this time the horizontal column givest.) - A stone is thrown at
15.0 m/sat35.0°above the horizontal from a cliff20.0 mhigh. Find its greatest height above the ground below, the time in the air, the horizontal distance, and the landing speed and angle. Confirm the speed with the angle-free check.
1. t = 0.404 s, x = 1.01 m. 2. T = 4.33 s, H = 23.0 m, R = 91.8 m. 3. Both give R = 900 × sin 30° ÷ 9.8 = 45.9 m, since sin 30° = sin 150°. 4. t = 25.0 ÷ 13.79 = 1.81 s, at which point y = 4.87 m, so it clears the fence by 1.87 m (already on its way down). 5. Greatest height 20.0 + 8.60² ÷ 19.6 = 23.8 m, t = 3.08 s, x = 37.9 m, v_y = −21.6 m/s, so the landing speed is √617 = 24.8 m/s at 60.4° below the horizontal.