Time of Flight Calculator – Projectile Motion

Enter initial velocity, launch angle, and launch height to find time of flight (T = 2v₀sin(θ)/g), maximum height, horizontal range, and impact speed.

Quick Facts

Time of flight
T = [v0sin(θ) + √((v0sin(θ))² + 2gh0)] / g
Reduces to T = 2v0sin(θ)/g when launch height h0 = 0.
Maximum height
H = h0 + (v0sin(θ))² / (2g)
Reached at t = v0sin(θ)/g, the midpoint of flight when h0 = 0.
Horizontal range
R = v0cos(θ) × T
Maximized at a 45° launch angle when launch and landing heights are equal.

Your Results

Calculated
Time of Flight
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T, total time airborne (s)
Maximum Height
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H, above the ground/landing plane
Horizontal Range
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R, horizontal distance traveled
Impact Speed
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Speed magnitude at landing

Ready

Enter initial velocity, launch angle, and launch height, then press Calculate.

About the Time of Flight Calculator – Projectile Motion

Time of flight is the total duration a projectile spends in the air, from the instant it launches until it reaches the landing height. For a launch and landing at the same height, the classic formula is T = 2v₀sin(θ)/g. This calculator uses the more general form that also accounts for a launch height above the landing point, so it works for problems like a ball kicked from a rooftop or a projectile fired from an elevated platform, not just launches starting at ground level.

Deriving the time-of-flight formula

Projectile motion splits into two independent components: horizontal motion at constant velocity vₓ = v₀cos(θ), and vertical motion under constant gravitational acceleration g, with initial vertical velocity vy₀ = v₀sin(θ). Starting from launch height h₀, the vertical position at time t is y(t) = h₀ + v₀sin(θ)t − ½gt². Setting y(t) = 0 and solving the resulting quadratic for t gives the general time of flight: T = [v₀sin(θ) + √((v₀sin(θ))² + 2gh₀)] / g. When h₀ = 0, the square root simplifies to v₀sin(θ) and the expression reduces to the familiar T = 2v₀sin(θ)/g. The same vertical component gives the maximum height H = h₀ + (v₀sin(θ))²/(2g), reached at t = v₀sin(θ)/g. Combined with the constant horizontal speed, this also gives the horizontal range R = vₓ × T and, from energy conservation, the impact speed at landing.

Practical notes and common mistakes

  • Angle convention: θ is measured from the horizontal, not the vertical. A 90° angle means straight up (zero horizontal range); a 0° angle means a flat, horizontal launch.
  • This model ignores air resistance. Drag reduces both range and time of flight, sometimes substantially for lightweight or high-speed projectiles such as golf balls, arrows, or bullets. Treat the result as an ideal upper bound.
  • Match gravity to your planet and unit system. Earth's standard gravity is 9.81 m/s² (32.2 ft/s²); the Moon is about 1.62 m/s² and Mars about 3.72 m/s². Mixing unit systems between velocity, height, and gravity is the most common source of wrong answers.
  • Launch height extends flight time, not the range formula directly — range is still vₓ × T, but T itself grows because gravity has farther to pull the projectile down before it reaches the landing plane.

Frequently Asked Questions

What is the formula for time of flight in projectile motion?
For a projectile launched and landing at the same height, time of flight is T = 2v₀sin(θ)/g. If the launch height h₀ differs from the landing height, the general formula is T = [v₀sin(θ) + √((v₀sin(θ))² + 2gh₀)] / g, where v₀ is initial speed, θ is the launch angle measured from horizontal, and g is gravitational acceleration.
Does time of flight depend on the mass of the projectile?
No. In the idealized model with no air resistance, mass cancels out of the equations of motion entirely — a bowling ball and a tennis ball launched with the same speed and angle from the same height stay airborne for the same amount of time. Air resistance, which this calculator ignores, is the main reason mass matters for real trajectories.
How does launch angle affect time of flight and range?
Time of flight increases with launch angle, from 0 seconds at 0° (flat launch) up to a maximum at 90° (straight up), because a steeper angle puts more of the initial speed into the vertical component. Range behaves differently: for equal launch and landing heights, range is maximized at 45°, and complementary angles such as 30° and 60° produce the same range but different times of flight.
Why can impact speed be greater than the initial launch speed?
When the launch height is above the landing point, the impact speed exceeds the initial speed because gravity does additional positive work on the projectile as it falls that extra distance. When launch and landing heights are equal, impact speed equals initial speed by conservation of energy.

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