Time of Flight Formula: T = 2u sin θ / g with Examples

The time of flight formula is $T = \frac{2u \sin\theta}{g}$ . It gives the total time a projectile spends in the air when it is launched at speed u and angle θ above the horizontal and lands back at the height it started from. Only the vertical part of the launch velocity, $u \sin\theta$ , decides how long the flight lasts.

In US books the launch speed is often written v₀, so the same result reads T = 2v₀ sin θ / g. This page uses u and g = 9.8 m/s², and ignores air resistance, as Class 11 and AP Physics 1 problems do unless they say otherwise.

Quick facts

Formula
$T = \frac{2u\sin\theta}{g}$
Time to the top
$t_{top} = \frac{u\sin\theta}{g} = \frac{T}{2}$
Horizontal launch from height h
$T = \sqrt{\frac{2h}{g}}$
SI unit
second (s)
Longest flight
θ = 90°, a straight up throw, giving T = 2u/g
Assumes
Level ground and no air resistance

Deriving the time of flight formula

Split the launch velocity into parts. The horizontal part is $u\cos\theta$ and stays the same all flight. The vertical part starts at $u\sin\theta$ and is pulled down by g. Take up as positive and write the vertical displacement:

$y = (u\sin\theta)\,t - \tfrac{1}{2} g t^2$

The projectile lands when y = 0 again. Factor out t: $t\left(u\sin\theta - \tfrac{1}{2} g t\right) = 0$ . One answer is t = 0, the launch. The other is the landing time:

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

The rise and the fall take equal times on level ground, so the top of the path is reached at exactly T/2.

Formula of time of flight: worked examples

Example 1: a football kick

A ball is kicked at 20 m/s at 30° above the horizontal. Find its time of flight.

$T = \frac{2 \times 20 \times \sin 30^\circ}{9.8} = \frac{2 \times 20 \times 0.5}{9.8} = \frac{20}{9.8} \approx 2.04\ \text{s}$

Example 2: launch speed in km/h

A stone leaves a catapult at 54 km/h at 60° to the ground. How long is it in the air?

54 km/h is 15 m/s, and sin 60° ≈ 0.866.

$T = \frac{2 \times 15 \times 0.866}{9.8} = \frac{25.98}{9.8} \approx 2.65\ \text{s}$

Example 3: rolled off a cliff

A ball rolls horizontally off a cliff 4.9 m high. How long does it take to hit the ground below?

There is no vertical launch speed, so $T = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 4.9}{9.8}} = \sqrt{1.0} = 1.0\ \text{s}$ .

The horizontal speed has no effect on this time.

Special cases

Time of flight in different launches
Launch	Time of flight	Why
Angle θ, level ground	$\frac{2u\sin\theta}{g}$	Rises and falls the same height
Straight up, θ = 90°	$\frac{2u}{g}$	sin 90° = 1, the longest flight for a given u
Horizontal from height h	$\sqrt{\frac{2h}{g}}$	Starts with zero vertical velocity
Angle θ from height h	Solve the quadratic for y = −h	Landing point is below launch point

Two launches at complementary angles, such as 30° and 60°, land at the same distance but do not take the same time. The steeper one stays up longer, because its vertical speed is larger.

How it links to range and height

Time of flight is the bridge between the vertical and horizontal parts of projectile motion. Multiply it by the horizontal speed and you get the range of projectile formula, R = u cos θ × T. Use half of it in the vertical equations and you reach the maximum height of projectile formula. Questions often ask for all three, so find T first.

The general formula of time page covers non projectile cases, and the projectile section of the kinematics formulas hub shows the full set.

Errors to watch for

Using cos instead of sin. The vertical part of the velocity uses sin θ when θ is measured from the horizontal.
Calculator in radians. sin 30 in radian mode gives −0.988. Set degree mode.
Using the formula when landing height differs. T = 2u sin θ / g quietly assumes the ball comes down to its starting height.
Forgetting the 2. u sin θ / g is only the time to the top.

Frequently asked questions

What is the formula for time of flight of a projectile?

T = 2u sin θ / g, where u is the launch speed, θ the angle above the horizontal and g = 9.8 m/s². It assumes the projectile lands at its launch height.

What is the time of flight for a horizontal projectile?

For a body launched horizontally from height h, T = √(2h / g). The horizontal speed does not change it.

At what angle is the time of flight greatest?

At 90°, a straight up throw, since sin θ is largest there. The flight time is then 2u / g.

Does mass affect time of flight?

No. Without air resistance, all objects fall with the same acceleration g, so mass does not appear in the formula.

How is time of flight related to time of ascent?

On level ground the time to reach the top is exactly half the time of flight, u sin θ / g.