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Projectile Motion Time Calculator

Time of flight, time to apex and hang time from launch speed, angle and height
A projectile is launched at 25 m/s at 40 degrees on level ground. Find the time of flight.
A ball is thrown horizontally at 12 m/s from a 45 m cliff. How long until it lands?
Find the time of flight for 20 m/s at 35 degrees launched from a 15 m cliff
How long does a projectile take to reach maximum height at 30 m/s and 60 degrees?

The Time of Flight Equation

Projectile motion splits into two independent problems. Horizontally the velocity never changes; vertically the object is in free fall. Time is the one quantity shared by both, which is why you almost always solve for it first.

The vertical equation is

y=y0+v0sin⁡θ t−12gt2y = y_0 + v_0 \sin\theta\, t - \tfrac{1}{2} g t^2

  • v0v_0 — launch speed, m/s; θ\theta — launch angle above the horizontal
  • gg — 9.81 m/s²; tt — time, s; y0y_0 — launch height, m

Level ground (lands at the launch height). Set y=y0y = y_0 and solve:

T=2v0sin⁡θgT = \frac{2 v_0 \sin\theta}{g}

The apex comes at exactly half of that, tup=v0sin⁡θ/gt_{\text{up}} = v_0\sin\theta / g.

When it applies: no air resistance, constant gg, and — for the symmetric formula — landing at the same height it left.

The assumption people forget: only the vertical component sets the time. Horizontal speed never appears in the time of flight.

Launching From a Height

If the projectile lands below its launch point, the flight is no longer symmetric and the symmetric formula overestimates nothing — it simply does not apply. Put y=0y = 0 and y0=hy_0 = h and solve the quadratic:

12gt2−v0sin⁡θ t−h=0\tfrac{1}{2} g t^2 - v_0\sin\theta\, t - h = 0

T=v0sin⁡θ+(v0sin⁡θ)2+2ghgT = \frac{v_0 \sin\theta + \sqrt{(v_0\sin\theta)^2 + 2gh}}{g}

with hh in metres. Take the positive root; the negative root is the time the projectile would have left the ground had it started there.

Two special cases fall straight out:

  • Horizontal launch (θ=0\theta = 0): T=2h/gT = \sqrt{2h/g}, independent of launch speed entirely.
  • Dropped from rest: the same expression, because a dropped object and a horizontally thrown one hit the ground together.

Once TT is known, the horizontal range is just R=v0cos⁡θ⋅TR = v_0\cos\theta \cdot T, in metres.

Common Mistakes to Avoid

  • Using the full launch speed as the vertical component — it is v0sin⁡θv_0\sin\theta, not v0v_0.
  • Applying T=2v0sin⁡θ/gT = 2v_0\sin\theta/g off a cliff — that formula assumes equal launch and landing heights. Use the quadratic instead.
  • Degrees versus radians — a calculator in radian mode turns sin⁡40°\sin 40° into sin⁡(40 rad)\sin(40\ \text{rad}) and the answer is silently wrong.
  • Sign confusion on gg — if you write −12gt2-\tfrac12 g t^2, then g=+9.81g = +9.81. Do not make it negative twice.
  • Reporting the time to the apex as the time of flight — on level ground it is half.
  • Assuming horizontal speed shortens the flight — a faster horizontal launch travels further in the same time, not for less time.

Examples

Step 1: Vertical component: v0y=25 m/s×sin⁡40°=25×0.6428=16.07 m/sv_{0y} = 25\ \text{m/s} \times \sin 40° = 25 \times 0.6428 = 16.07\ \text{m/s}
Step 2: T=2v0y/g=(2×16.07 m/s)÷(9.81 m/s2)T = 2v_{0y}/g = (2 \times 16.07\ \text{m/s}) \div (9.81\ \text{m/s}^2)
Step 3: T=32.14÷9.81=3.276 sT = 32.14 \div 9.81 = 3.276\ \text{s}
Step 4: Horizontal component: v0x=25×cos⁡40°=19.15 m/sv_{0x} = 25 \times \cos 40° = 19.15\ \text{m/s}
Step 5: R=v0xT=(19.15 m/s)(3.276 s)=62.7 mR = v_{0x} T = (19.15\ \text{m/s})(3.276\ \text{s}) = 62.7\ \text{m}
Answer: T≈3.28T \approx 3.28 s, range ≈62.7\approx 62.7 m

Step 1: Horizontal launch, so v0y=0v_{0y} = 0 and h=12gT2h = \tfrac{1}{2} g T^2
Step 2: T=2h/g=(2×45 m)÷(9.81 m/s2)T = \sqrt{2h/g} = \sqrt{(2 \times 45\ \text{m}) \div (9.81\ \text{m/s}^2)}
Step 3: T=9.174 s2=3.029 sT = \sqrt{9.174\ \text{s}^2} = 3.029\ \text{s}
Step 4: Horizontal distance: x=(12 m/s)(3.029 s)=36.3 mx = (12\ \text{m/s})(3.029\ \text{s}) = 36.3\ \text{m}
Answer: T≈3.03T \approx 3.03 s, landing ≈36.3\approx 36.3 m from the base

Step 1: v0y=20 m/s×sin⁡35°=20×0.5736=11.47 m/sv_{0y} = 20\ \text{m/s} \times \sin 35° = 20 \times 0.5736 = 11.47\ \text{m/s}
Step 2: T=v0y+v0y2+2ghgT = \dfrac{v_{0y} + \sqrt{v_{0y}^2 + 2gh}}{g}
Step 3: v0y2=(11.47 m/s)2=131.6 m2/s2v_{0y}^2 = (11.47\ \text{m/s})^2 = 131.6\ \text{m}^2\text{/s}^2; 2gh=2(9.81)(15)=294.3 m2/s22gh = 2(9.81)(15) = 294.3\ \text{m}^2\text{/s}^2
Step 4: 131.6+294.3=425.9=20.64 m/s\sqrt{131.6 + 294.3} = \sqrt{425.9} = 20.64\ \text{m/s}
Step 5: T=(11.47+20.64) m/s÷9.81 m/s2=32.11÷9.81=3.273 sT = (11.47 + 20.64)\ \text{m/s} \div 9.81\ \text{m/s}^2 = 32.11 \div 9.81 = 3.273\ \text{s}
Answer: T≈3.27T \approx 3.27 s (the level-ground formula would have given only 2.342.34 s)

Frequently Asked Questions

For a launch and landing at the same height, T = 2v₀sinθ/g. Only the vertical component of the launch velocity matters, so a projectile fired at 25 m/s and 40° stays airborne for 3.28 s regardless of how far it travels horizontally.

Solve the vertical quadratic ½gt² − v₀sinθ·t − h = 0 and keep the positive root: T = [v₀sinθ + √((v₀sinθ)² + 2gh)]/g, with h the launch height in metres. The symmetric formula is only valid for equal launch and landing heights.

No. Gravity acts only vertically, so the horizontal velocity has no effect on how long the projectile is in the air. A ball thrown horizontally at 12 m/s and one simply dropped from the same height hit the ground at the same instant.

t = v₀sinθ/g, the moment the vertical velocity passes through zero. On level ground that is exactly half the total time of flight, because the rise and the fall are mirror images.

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