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  1. Home
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  3. /Classical Mechanics - Kinematics Calculators
  4. /Time of Flight Calculator

Time of Flight Calculator

Last updated: March 17, 2026

Calculator

Results

Total Time of Flight

—

s

Time to Maximum Height

—

s

Horizontal Range

—

m

Maximum Height

—

m

Impact Speed

25

m/s

Results

Total Time of Flight

—

s

Time to Maximum Height

—

s

Horizontal Range

—

m

Maximum Height

—

m

Impact Speed

25

m/s

The Time of Flight Calculator determines how long a projectile remains airborne from launch to landing on level ground. Time of flight is a fundamental quantity in projectile motion that directly influences the horizontal range and is essential for timing-dependent applications such as sports strategy, fireworks choreography, and ballistic planning.

For a projectile launched and landing at the same height, the time of flight is: $$T = \frac{2v_0 \sin(\theta)}{g}$$ where $$v_0$$ is the initial speed, $$\theta$$ is the launch angle above the horizontal, and $$g$$ is gravitational acceleration. This formula follows directly from the vertical motion equation by finding when the projectile returns to its launch height.

The trajectory is perfectly symmetric about its midpoint: the time to reach maximum height equals exactly half the total flight time, $$t_H = T/2 = v_0 \sin\theta / g$$. At this moment, the vertical velocity is zero and the projectile transitions from rising to falling. This symmetry is a direct consequence of constant gravitational acceleration and the absence of air resistance.

Time of flight increases linearly with the vertical velocity component $$v_0 \sin\theta$$. Launching at 90° (vertically) maximizes flight time to $$T_{\max} = 2v_0/g$$, while a horizontal launch (0°) yields zero flight time (the object is already at ground level). Doubling the initial speed doubles the flight time, unlike range which quadruples.

In real-world applications, air resistance breaks the perfect symmetry: the descent takes longer than the ascent because drag decelerates the projectile going up but only partially counteracts gravity coming down. For a baseball, actual flight time might be 5-15% longer than the ideal calculation predicts, depending on speed and spin.

The impact speed for level-ground projectile motion equals the launch speed ($$v_0$$) in the ideal case, a consequence of energy conservation: the projectile returns to its launch height with the same kinetic energy. The velocity components at landing are the same magnitudes as at launch, but the vertical component is reversed.

Visual Analysis

How It Works

Enter the initial velocity and launch angle. The calculator applies $$T = 2v_0 \sin\theta / g$$ for total flight time and divides by 2 for the ascent time. Range is computed as $$R = v_0^2 \sin(2\theta)/g$$ and maximum height as $$H = v_0^2 \sin^2\theta/(2g)$$. Impact speed equals launch speed by energy conservation on level ground.

Understanding Your Results

Longer flight time means the projectile spends more time airborne but not necessarily greater range—a steep angle gives long flight time with short range. The impact speed equaling launch speed confirms energy conservation. For timing applications (e.g., catching a ball), the half-time marks the apex moment.

Worked Examples

Baseball outfield throw at 45°

Inputs

v025
angle45
g9.81

Results

total time3.604
half time1.802
range val63.71
max height15.928
impact speed25

A 25 m/s throw at 45° stays airborne 3.6 s, reaches 15.9 m high, and travels 63.7 m.

High-angle mortar at 70°

Inputs

v040
angle70
g9.81

Results

total time7.66
half time3.83
range val104.678
max height71.935
impact speed40

At 70°, the projectile is airborne 7.7 s—much longer than at 45° (5.8 s)—but range is less.

Frequently Asked Questions

90° (vertical launch) maximizes flight time to $$2v_0/g$$, since all velocity is directed upward. However, the horizontal range is zero at this angle.

Yes, in the ideal case (no air resistance, level ground). The trajectory is perfectly symmetric. With air drag, descent takes slightly longer than ascent.

By conservation of energy: the projectile returns to the same height with the same potential energy, so kinetic energy (and thus speed) is unchanged. Only the direction of the vertical component reverses.

Time of flight is inversely proportional to $$g$$. On the Moon ($$g = 1.62\,\text{m/s}^2$$), a projectile stays airborne about 6 times longer than on Earth with identical launch conditions.

No, this formula assumes level ground (launch and landing at the same height). For elevated launches, you need the full quadratic equation involving the launch height $$h$$.

Range equals horizontal velocity times flight time: $$R = v_0 \cos\theta \times T$$. Longer flight time increases range only if horizontal velocity is sufficient—the optimal balance occurs at 45°.

Sources & Methodology

Halliday, Resnick & Walker, Fundamentals of Physics (11th ed., Wiley, 2018); Young & Freedman, University Physics (15th ed., Pearson, 2020); Serway & Jewett, Physics for Scientists and Engineers (10th ed., Cengage, 2019).
R

Roboculator Team

The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.

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