Projectile Motion Time Calculator

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

$y = y_0 + v_0 \sin\theta\, t - \tfrac{1}{2} g t^2$

• $v_0$ — launch speed, m/s; $\theta$ — launch angle above the horizontal
• $g$ — 9.81 m/s²; $t$ — time, s; $y_0$ — launch height, m

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

$T = \frac{2 v_0 \sin\theta}{g}$

The apex comes at exactly half of that, $t_{\text{up}} = v_0\sin\theta / g$.

When it applies: no air resistance, constant $g$, 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.

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 $y = 0$ and $y_0 = h$ and solve the quadratic:

$\tfrac{1}{2} g t^2 - v_0\sin\theta\, t - h = 0$

$T = \frac{v_0 \sin\theta + \sqrt{(v_0\sin\theta)^2 + 2gh}}{g}$

with $h$ 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 ($\theta = 0$): $T = \sqrt{2h/g}$, 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 $T$ is known, the horizontal range is just $R = v_0\cos\theta \cdot T$, in metres.

• Using the full launch speed as the vertical component — it is $v_0\sin\theta$, not $v_0$.
• Applying $T = 2v_0\sin\theta/g$ off a cliff — that formula assumes equal launch and landing heights. Use the quadratic instead.
• Degrees versus radians — a calculator in radian mode turns $\sin 40°$ into $\sin(40\ \text{rad})$ and the answer is silently wrong.
• Sign confusion on $g$ — if you write $-\tfrac12 g t^2$, then $g = +9.81$. 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.