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Projectile Position at Time Calculator

Enter the initial velocity, launch angle, and time to find the projectile's position.

How it is calculated

The tool places a projectile at time t using x = v0 · cos(θ) · t and y = v0 · sin(θ) · t − ½ · g · t2. Speed v0 is in m/s, the launch angle θ in degrees above the horizontal, time in seconds, and gravity g in m/s². If you leave g empty it uses 9.80665. Results are in metres, measured from the launch point, with y positive upward. A note says whether the projectile is at or above the launch height, or below it. Speed, time and g must all be above zero. The model has no air resistance, and it has no ground: for a long time, y keeps going negative as if the projectile fell through the floor.

Worked example

A ball is launched at 25 m/s at 40 degrees, and you want its position after 2 s. Horizontally, 25 × cos 40° × 2 = 25 × 0.7660 × 2 = 38.30 m. Vertically, 25 × sin 40° × 2 = 32.14 m of climb, minus the fall under gravity ½ × 9.80665 × 22 = 19.61 m. That leaves y = 12.53 m. The tool shows 38.30 m and 12.53 m with the note that it is at or above launch height. At 4 s the same launch gives y = −14.17 m, which means it would be below the launch point.

Try this example →

Common mistakes

  • Entering the angle in radians. The field is degrees, so 0.7 would be read as 0.7 degrees, not about 40.
  • Entering a time after the projectile would have hit the ground and reading a negative y as a real position. The tool has no ground level, so a negative y simply means below the launch height.
  • Entering the speed in km/h. It must be in m/s, so divide km/h by 3.6.
  • Expecting air resistance to be included. The tool ignores it, so real objects, especially light or fast ones, will land short of this prediction.
  • Using the launch height as the ground height when the launch point is raised, such as from a cliff. All positions are relative to the launch point.

About this calculator

A projectile launched at an angle follows two independent motions at once: constant horizontal velocity (ignoring air resistance) and vertically accelerated motion under gravity. Its position at any time t is given by x = v₀·cos(θ)·t for the horizontal distance and y = v₀·sin(θ)·t − ½g·t² for the height, where v₀ is the initial speed, θ the launch angle, and g gravitational acceleration (9.8 m/s²).

This calculator plugs your initial velocity, launch angle, and a specific time into those two equations to find exactly where the projectile is — not just at the peak or landing point, but at any moment during the flight. It's a core tool in introductory physics courses for checking projectile motion homework and visualizing how horizontal and vertical motion combine independently.

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Comments (1)

  • Editorial team

    To find where a throw comes back down to its launch height, try times until y reads close to zero. For the default 20 m/s at 45 degrees that happens near 2.88 seconds, with x at about 40.8 metres. Past that moment y keeps going negative, because the tool has no ground. If you are checking a physics problem, post the numbers you were given and the answer in your book so others can compare.

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