Article
You stand on a bridge edge and push a ball sideways. Vertically it falls like a dropped object; horizontally it keeps a steady speed. Below you will see how to find flight time and range.
Part 5 of Physics without mysteries. Compare throws: vertical, horizontal and angled side by side.
On this page: v₀ is the horizontal speed at launch. h is the launch height. Z is the range, the horizontal distance. g is gravitational acceleration, about 9.81 m/s². vx stays constant because there is no horizontal acceleration.
You stand on the edge of a bridge or roof and push a ball sideways. You give it no upward or downward speed: you start from height h with speed only in the horizontal direction. That is a horizontal throw.
At launch vy0 = 0, so the vertical motion is free fall from that height. Horizontally, with no drag, vx = v0 stays constant. Flight time comes from the fall. Range comes from constant vx.
The vector split is simple: vx = v0, vy0 = 0, ax = 0, ay = -g (y upward). With y = 0 at the landing level and launch at y = h:
x(t) = v0·t
y(t) = h - ½·g·t²
The path in the xy plane is a parabolic arc. Weight has no horizontal component in this model, so x motion is uniform and y motion is independent. Close t from h and g first, then multiply by v0.
Landing y(t) = 0 gives ½·g·t² = h, so the flight time is:
t = √(2h/g)
Horizontal range is the x distance in that time:
Z = v0·t = v0·√(2h/g)
Units: h in meters, v0 in m/s, t in seconds, Z in meters, g in m/s². If your speed is in km/h, divide by 3.6. First intuition, a bridge: h = 20 m, v0 = 10 m/s, g = 9.81 m/s² gives t ≈ 2.02 s and Z ≈ 20.2 m. That is the same set you enter in the horizontal-throw calculator.Units: h in feet, v0 in ft/s, t in seconds, Z in feet, g in ft/s². If your speed is in mph, convert to ft/s. First intuition, a bridge: h = 20 m, v0 = 10 m/s, g = 9.81 m/s² gives t ≈ 2.02 s and Z ≈ 20.2 m. That is the same set you enter in the horizontal-throw calculator.
A taller bridge at the same v0 means longer flight and larger Z, but the dependence on h is a square root, not linear. Two balls from the same height, one dropped and one pushed horizontally, hit the water at the same instant in the no-drag model. The second one is only farther in x.
At landing, vx = v0 never changes. The y component grows as in free fall: |vy| = √(2·g·h). The speed magnitude is v = √(v0² + 2·g·h). Many problems only need Z and t; landing speed appears when they ask for energy or how hard it hits.
If the launch angle is not zero, it is already projectile motion, even with h0 > 0. Do not force cos α ≈ 1 when the problem gives α = 10° or 15°. Conversely, a horizontal throw is the angled-throw limit as α → 0° from a tower. If h → 0 at α = 0, range vanishes: there is no flight time.
A common mistake sounds like this: “range is v0²/g, because that is what we used at 45°.” That formula belongs to flat-ground projectile motion, not to a bridge. Here there is no 45° angle and no h0 = 0. Take time from the fall √(2h/g), then Z = v0·t. Another trap is t = h/v0, as if the vertical motion were uniform.
Second numeric intuition: if you are given v in km/h, convert at once. With h = 45 m and v0 = 72 km/h (that is 20 m/s), time t = √(90/9.81) ≈ 3.03 s and Z ≈ 60.6 m. Compute that example once on paper and once in the calculator to see where zeros usually disappear in unit conversion.Second numeric intuition: if you are given v in mph, convert at once to ft/s. With h = 45 m and v0 = 72 km/h (that is 20 m/s), time t = √(90/9.81) ≈ 3.03 s and Z ≈ 60.6 m. Compute that example once on paper and once in the calculator to see where zeros usually disappear in unit conversion.
At launch Em = ½·m·v0² + m·g·h (if Ep = 0 on the ground). At landing Ep = 0 and Ek = ½·m·(v0² + 2·g·h). With no drag, energy is conserved; you still compute range from kinematics, because energy alone does not say how far you travel in x.
When the problem gives flight time instead of h, invert the formula: h = ½·g·t², then Z = v0·t. When it gives Z and h, you can find v0 = Z / √(2h/g). The algebra runs both ways, even if the calculator is built for the usual inputs v0, h, g.
Watch the effective height: if you throw from a hand above the edge, h is the vertical distance from the release point to the landing level, not the bridge height in a datasheet. A few tens of centimeters at small h change t a lot, because t depends on the square root of h.
One more unit trap: mixing “stories” with meters or feet without converting. If the problem says “from the third floor”, turn that into a concrete h before you take the square root. Otherwise, Z becomes guesswork.
Switch to full projectile motion when you are given an angle, you want H_max above the start, a path chart, or landing is not the simple “flat ground under a tower” case. Then use full projectile motion. Keep the horizontal tool when the problem says horizontal throw or “pushed horizontally”.
Before you calculate, check: vy0 = 0? h known in length units? g fixed? t from the vertical? Z from v0·t? If any point fails, you are not in a pure horizontal throw or you lack data. Then return to the throw comparison or to projectile motion.
Horizontal throw trains one habit: split the components. Vertical sets time. Horizontal sets range. Once that clicks, moving to an angled throw is mostly adding vy0 = v0·sin α and a longer equation for t.
Open horizontal throw. Enter v0, height h, and g = 9.81 m/s². Read flight time t and range Z.
If the problem gives an angle, or you want the full path and landing components, do not force this tool. Switch to full projectile motion.
Launch is exactly horizontal (α = 0), so vy0 = 0. Constant g, no drag, landing at y = 0 after a fall of h. We work in meters, seconds and m/s.Launch is exactly horizontal (α = 0), so vy0 = 0. Constant g, no drag, landing at y = 0 after a fall of h. We work in feet, seconds and ft/s.
Three concrete cases: (1) α ≠ 0, so vy0 is not zero, (2) you want H_max above the start or the full path, (3) h = 0 and range vanishes because there is no flight time. Then use projectile motion and the full model instead of forcing √(2h/g).
| Quantity | Formula | Source |
|---|---|---|
| Flight time | t = √(2h/g) | Vertical fall with vᵧ₀ = 0 |
| Range | Z = v₀·t | Constant vₓ = v₀ |
| Path | y = h - ½gt², x = v₀t | Parabola in the plane |
| α = 0, h₀ > 0 | full projectile model | When you also want landing vᵧ |
From a bridge of height h = 20 m you throw horizontally at v0 = 10 m/s. Take g = 9.81 m/s². Find flight time and range.
Steps
Answer: t ≈ 2.02 s, Z ≈ 20.2 m
The vertical part starts with vy0 = 0, so it behaves like ordinary free fall from height h. That gives t = √(2h/g), where h is launch height above the landing level.
Find time from the fall first, then multiply by constant vx. Z = v0·√(2h/g) is about 20.2 m with g = 9.81 m/s².
In the no-drag model no: ax = 0, so vx stays constant. Gravity accelerates only vertically, and that is what sets the flight time.
Horizontal means α = 0 and vy0 = 0, so time comes from the fall alone. Angled throws have a nonzero vertical launch component and usually sin/cos formulas.
Not for a pure horizontal throw. That formula is for flat-ground projectile motion (h0 = 0) at angle α, not for a bridge launch at α = 0.
Nothing slows the horizontal part, while the vertical part gains speed as in free fall. So v = √(v0² + 2·g·h), with components vx = v0 and |vy| = √(2·g·h).
Even a small angle adds vy0, so treat it as projectile motion with the given α and h0. Do not force √(2h/g) without checking the vertical component.
Usually g = 9.81 m/s². If the problem says 10, use 10. Keep h and v0 consistent with the unit choice in the page header.
When the problem is about a horizontal push from height h, use horizontal throw. When an angle appears or you want the full path, switch to full projectile motion.
In reality yes; it usually shortens range and slightly affects t. In a typical problem we take t = √(2h/g) with no drag and treat the result as an estimate.