Example 1
v₀ = 15 m/s, upward, g = 9.81 m/s².
Pick throw up or throw down and enter v₀. On the way up the calculator computes H_max = v₀²/(2g). 15 m/s is about 11.5 m and 1.53 s to the peak, vertical only, no range Z.
Throw direction, Initial speed v₀, Time t (optional) and g (default 9.81). The result shows up here.
A vertical throw moves only on the vertical. Upward: v(t) = v₀ − g t, h(t) = v₀ t − ½ g t². At the peak v = 0, so H_max = v₀²/(2g) and climb time is v₀/g. 15 m/s is H_max ≈ 11.5 m and about 1.53 s to the peak. Return to launch level takes twice that, about 3.06 s. At 10 m/s after 1 s: v ≈ 0.19 m/s, h ≈ 5.1 m.
Fields: direction, v₀, optional t, empty g = 9.81. Downward, H_max is not a climb peak: you already start with speed down. Blank t means you want H_max and the characteristic times, not the state at time t. A comma and a period are the same v₀: 12,5 and 12.5.
v₀ must be a number. Zero m/s up is no throw. A time longer than the return shows the formula state; ground impact at h = 0 uses flight time 2 v₀/g when launch was at ground level.
A drop from v₀ = 0 lives on the free-fall page: there you type t and get h. A nonzero angle is on full projectile. This calculator has no Z, because there is no horizontal component.
v₀ is squared in H_max. Twice the speed is four times the peak, not two. 30 m/s up is about 45.9 m, not 23 m.
Pick up, type 15, leave t empty, click Calculate, and match H_max ≈ 11.5 m. Then add t = 1 at v₀ = 10 and compare v and h. The header symbol does not toss water.
Drop from rest: free fall. Arc with an angle: full projectile.
Upward throw (axis upward):
v(t) = v0 − g t
h(t) = v0t − ½g t2
Hmax = v02 / (2g), time to apex tup = v0/g, round-trip flight time 2tup.
A vertical throw has no range Z. 15 m/s upward is H_max ≈ 11.5 m and about 1.53 s to the apex; return to launch level takes twice that.
v₀ = 15 m/s, upward, g = 9.81 m/s².
v₀ = 10 m/s, t = 1 s: where is it after one second?
v₀ = 8 m/s almost straight up.
Water leaves the nozzle upward at 12 m/s.
v₀ = 5 m/s downward, t = 1.5 s.
v₀ = 15 m/s upward, g = 1.62 m/s².
Takeoff speed 4 m/s upward: center-of-mass H_max.
v₀ = 20 m/s upward: full flight time.
H_max = 225 / 19.62 ≈ 11.5 m. Time to peak ≈ 1.53 s. Return to launch level ≈ 3.06 s.
No. t is optional. Without t you get H_max and the characteristic times. With t you get v(t) and h(t).
v₀ in m/s, t in seconds, g in m/s². Result in meters. Empty g = 9.81.
Up, that is no throw. For a drop from rest, open free fall: there t gives h = ½ g t².
Here you have a downward v₀. Free fall starts at v₀ = 0 and computes h from t. Different unknown, different card.
This calculator has no Z, because there is no horizontal component. Open horizontal throw or full projectile.
No. Classroom vacuum model. In air the peak sits lower than 11.5 m at 15 m/s.
The result is the formula state. Ground impact at h = 0 uses flight time 2 v₀/g when launch was at ground level.
v ≈ 0.19 m/s, h ≈ 5.1 m. Still before the peak: climb time at 10 m/s is about 1.02 s.
Yes. 12,5 and 12.5 are the same v₀. The calculator does not require a period.
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.