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    Kinematics / trajectory

    Projectile motion

    Solve a launch into range, apex height, flight time and impact speed. The trajectory is sampled at ten points so the shape is explicit, and the 45° rule is shown to break down as soon as launch height is non-zero.

    Launch angle45.0°Horizontal speed17.68 m/sIdeal angle45°
    Range
    63.73 m
    Horizontal distance to impact
    Maximum height
    15.93 m
    Reached at 1.80 s
    Time of flight
    3.61 s
    Launch to ground contact
    Impact speed
    25.00 m/s
    At 45.0° below horizontal
    Trajectory sample points
    Position at ten equal time steps from launch to impact.
    63.7 m range
    Trajectory sample points
    Time (s)Horizontal (m)Height (m)Vertical speed (m/s)
    0.00 s0.000.0017.68
    0.36 s6.375.7414.14
    0.72 s12.7510.2010.61
    1.08 s19.1213.387.07
    1.44 s25.4915.303.54
    1.80 s31.8715.930.00
    2.16 s38.2415.30-3.54
    2.52 s44.6113.38-7.07
    2.88 s50.9910.20-10.61
    3.24 s57.365.74-14.14
    3.61 s63.730.00-17.68

    Vertical speed falls linearly with time; horizontal speed is constant because drag is ignored.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Trajectory
    Height against horizontal distance for the selected launch.
    Trajectory0.0003.577.1410.7114.2817.85005511111616212127273232373742424848535358586464Height (m)
    Range by launch angle
    Same launch speed, five angles — 45° is only optimal for h₀ = 0.
    Range by launch angle0.00015.0430.0845.1260.1675.2031.8715°15°55.1930°30°63.7345°45°55.1960°60°31.8775°75°Range (m)
    Canonical equations
    The exact relationships implemented by this engine.

    x = v₀cosθ·t · constant velocity in vacuum

    y = h₀ + v₀sinθ·t − ½gt² · constant acceleration

    t = [v_y + √(v_y²+2gh₀)]/g · solve y = 0

    R = v₀cosθ·t · horizontal distance at impact

    engineering methodology

    A clear path from inputs to results.

    In a vacuum, a projectile’s horizontal and vertical motions are independent: constant velocity sideways, constant acceleration downward. That decomposition gives closed-form answers for range, apex and flight time — and shows exactly why 45° is optimal only when launch and landing heights match.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Enter the launch speed.
    2. Set the launch angle above horizontal.
    3. Enter the launch height if the projectile starts above the landing plane.
    4. Adjust gravity for other planets or a different model.
    5. Read the range, apex and flight time.
    6. Compare angles in the range chart — the optimum shifts below 45° as launch height rises.

    The formula

    x=v0cos⁡θ t,y=h0+v0sin⁡θ t−12gt2,tf=vy+vy2+2gh0g,R=v0cos⁡θ tfx = v_0\cos\theta\, t, \quad y = h_0 + v_0\sin\theta\, t - \tfrac{1}{2}gt^2, \quad t_f = \frac{v_y + \sqrt{v_y^2 + 2gh_0}}{g}, \quad R = v_0\cos\theta\, t_f

    Reference: Standard vacuum ballistics; Newtonian kinematics.

    Worked example

    25 m/s at 45° from ground level (default inputs)

    Inputs
    • Launch speed 25 m/s.
    • Angle 45°.
    • Launch height 0 m.
    • Gravity 9.807 m/s².
    Calculation path
    1. 1

      Components: v_x = 17.68 m/s, v_y = 17.68 m/s.

    2. 2

      Flight time: 2 × 17.68 / 9.807 = 3.605 s.

    3. 3

      Range: 17.68 × 3.605 = 63.73 m.

    4. 4

      Apex: 17.68² / (2 × 9.807) = 15.93 m.

    5. 5

      Impact speed equals launch speed at ground level: 25 m/s.

    Result

    Range 63.73 m · apex 15.93 m · flight 3.61 s · impact 25 m/s.

    Interpreting the result

    At ground level the trajectory is symmetric: launch and impact speeds are equal and the apex sits at half the flight time. Launch from height and the symmetry breaks — the projectile spends longer falling than rising, the impact speed exceeds the launch speed, and the optimal angle drops below 45°. The angle chart makes that shift visible.

    Assumptions & limitations

    • Vacuum: no air resistance, no wind, no spin, no lift.
    • Constant gravity, which is accurate over ranges of a few kilometres.
    • Flat landing plane at the specified height.
    • Point mass with no size or orientation.
    • No atmospheric density variation with altitude.

    Frequently asked questions

    How much does air resistance change the answer?

    Substantially for fast or light projectiles. A golf ball loses about half its vacuum range; a dense artillery shell loses far less. Real trajectories are asymmetric — steeper descent, shorter range.

    How does launch height change the optimal angle?

    It lowers it. Launching from height gives the projectile free fall time, so more of the speed can go horizontal. The engine reports the optimum for the current inputs.

    What is the impact angle?

    The angle of the velocity vector at impact, below horizontal. It equals the launch angle for a symmetric trajectory and steepens when launching from height.

    Why is 45° optimal?

    Because range is proportional to sin(2θ), which peaks at 90°, so θ = 45°. This holds only when launch and landing heights are equal and there is no drag.

    Does mass matter?

    Not in a vacuum — all masses follow the same trajectory under gravity. In air, mass matters enormously because drag acceleration depends on the mass-to-area ratio.

    Can I model a different planet?

    Yes — set gravity. The Moon is 1.62 m/s², Mars 3.72, Jupiter 24.79. Range scales inversely with gravity.

    Methodology last reviewed 2026-10-18. Results are informational and not financial or engineering advice — see our disclaimer.