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Projectile Motion Calculator

Find a projectile's range, time of flight, maximum height and impact speed and angle from its launch speed, angle and height, with a table of positions.

Level: Grades 11–12

  • Common Core standard HSA.REI.B.4.bSolve quadratic equations by inspection (e.g., for x² = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b.
  • Common Core standard HSG.SRT.C.8Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems.
Printable worksheet

Projectile Motion Calculator

Know the answer you need? Pick an input to work backward.

Inputs

Measured from the horizontal: 0° is level, 90° straight up, negative angles point downward (−90° straight down).

0 when it lands at the same level it was launched from.

The flight time is split into this many equal steps for the position table.

Result

Range (horizontal distance)

40.7886 m

Range 40.7886 m, flight time 2.8842 s, peak height 10.1972 m.

Time of flight
2.8842 s
Maximum height above landing level
10.1972 m
Time to reach maximum height
1.4421 s
Impact speed
20 m/s
Impact angle below horizontal
45 °
Horizontal velocity (constant)
14.1421 m/s
Initial vertical velocity
14.1421 m/s

See the step-by-step solution ↓

Position every 0.2884 s (y is height above the landing level)
Position every 0.2884 s (y is height above the landing level)
Time (s)x (m)y (m)Vertical velocity (m/s)
00014.142
0.2884.0793.67111.314
0.5778.1586.5268.485
0.86512.2378.5665.657
1.15416.3159.7892.828
1.44220.39410.197-1.776e-15
1.73124.4739.789-2.828
2.01928.5528.566-5.657
2.30732.6316.526-8.485
2.59636.713.671-11.314
2.88440.7890-14.142

Step-by-step solution

  1. Use the surface gravity of Earth (standard gravity)

    g=9.80665 m/s²

  2. Find the sideways part of the launch speed, which stays the same for the whole flight: vₓ = v₀ × cos θ

    20×cos(45 °)=14.142136 m/s

  3. Find the upward part of the launch speed, which gravity works against: v_y = v₀ × sin θ

    20×sin(45 °)=14.142136 m/s

  4. Find the time in the air, the moment the height returns to the landing level: t = (v_y + √(v_y² + 2 g h₀)) ÷ g

    14.142136+14.1421362+2×9.80665×09.80665=2.884193 s

  5. Find the highest point, reached when the upward speed falls to zero: H = h₀ + v_y² ÷ (2g)

    0+14.14213622×9.80665=10.197162 m

  6. Find the range: the sideways speed never changes, so multiply it by the time in the air, R = vₓ × t

    14.142136×2.884193=40.788649 m

  7. Find the up-and-down speed at landing, since gravity has been slowing then speeding it the whole time: v_y − g × t

    14.142136−9.80665×2.884193=−14.142136 m/s

  8. Combine the sideways and up-and-down speeds to get the total speed at impact: √(vₓ² + v_y²)

    14.1421362+(−14.142136)2=20 m/s

  9. Find how steeply it comes down, measured below the horizontal: arctan(|v_y| ÷ vₓ)

    arctan(14.14213614.142136)=45 °

Test yourself

20 quick questions built from this calculator's formula: work out results, predict what happens when an input changes, and work backward from an answer. Every answer comes with the working.

What this calculator does

Find out how far, how high and how long something flies when you throw or launch it. A projectile is anything launched into the air and then left to gravity: a thrown ball, a jet of water, a long jumper. Without air resistance its path is a smooth arc called a parabola. The sideways speed never changes. The up-and-down speed drops by g (the pull of gravity, 9.80665 m/s² on Earth) every second. That first slows the rise and then speeds the fall.

Give this calculator the launch speed, the launch angle and how high above the landing level the launch happens. It reports how far the projectile travels, how long it stays in the air, how high it climbs, and how fast and how steeply it comes down. A table lists the x and y position at equal time intervals so you can plot the trajectory.

Where it's used

Why this math matters outside the classroom:

  • Kicked ball. A ball kicked from the ground at 20 m/s and 45° travels about 20² ÷ 9.81 ≈ 40.8 m, ignoring air resistance.
  • Best launch angle. Students find that, without air resistance, 45° gives the longest range for a launch from ground level.
  • Fountains. Fountain designers set the launch speed and angle of each water jet so its arc lands in the basin.
  • Firefighting. Firefighters angle a hose stream upward to reach a higher window or throw water farther.
  • Basketball shots. Coaches use launch angle and speed to explain why higher-arcing shots come down on a larger opening at the rim.

Grade level and standards: Grades 11–12; Common Core State Standards for Mathematics HSA.REI.B.4.b, HSG.SRT.C.8. © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved.

How to use it

  1. Enter the launch speed and choose its unit.
  2. Enter the launch angle in degrees from the horizontal, from −90° (straight down) to 90° (straight up).
  3. Enter the launch height above the landing level (0 for level ground) and choose the gravity.
  4. Choose the distance unit and the number of table intervals, then read the results, working and trajectory table.

What the inputs mean

Launch angle (θ)
Angles above 0° launch upward and below 0° launch downward. Values outside −90° to 90° are rejected, since they would point backward.
Launch height above the landing level (h₀)
Measure from the launch point down to where the projectile lands. A ball thrown from a 1.8 m shoulder height onto flat ground has h₀ = 1.8 m.
Custom gravitational acceleration
For example 9.81 or 9.8 m/s² to match a textbook, or 32.174 ft/s².
Trajectory table intervals
The table divides the flight time into this many equal steps, so 10 intervals give 11 rows from launch to landing.

Formula and method

vₓ = v₀ × cos θ,   v_y = v₀ × sin θ
t = (v_y + √(v_y² + 2 × g × h₀)) ÷ g
R = vₓ × t
H = h₀ + v_y² ÷ (2 × g)   (when v_y > 0)
x(t) = vₓ × t,   y(t) = h₀ + v_y × t − ½ × g × t²

Here v₀ is the launch speed and θ (theta) is the launch angle. vₓ and v_y are the sideways and upward parts of the speed, g is gravity and h₀ is the launch height. t is the time in the air, R is the range (distance traveled) and H is the highest point. x(t) and y(t) are the sideways distance and the height at time t.

The launch speed is split into a sideways part, which stays constant, and an upward part, which changes at the rate g. The time of flight is the moment the height y(t) comes back to zero. The range is then the sideways speed multiplied by that time.

At the peak the upward speed is zero, which gives the maximum height. At landing the vertical speed is v_y − g × t. Combining it with vₓ using the Pythagorean theorem gives the impact speed. The arctangent, arctan(|v_y| ÷ vₓ), gives the impact angle below the horizontal. On level ground (h₀ = 0) the range simplifies to v₀² × sin 2θ ÷ g, which is largest at 45°.

Worked example

A ball is thrown at 15 m/s at 35° above the horizontal from a height of 1.8 m on Earth, landing on flat ground. How far does it go?

  1. Components: vₓ = 15 × cos 35° = 12.287281 m/s; v_y = 15 × sin 35° = 8.603647 m/s.
  2. Time of flight: t = (8.603647 + √(8.603647² + 2 × 9.80665 × 1.8)) ÷ 9.80665 = 1.943537 s.
  3. Maximum height: H = 1.8 + 8.603647² ÷ (2 × 9.80665) = 5.574109 m.
  4. Range: R = 12.287281 × 1.943537 = 23.880783 m.
  5. Landing: v_y = 8.603647 − 9.80665 × 1.943537 = −10.45594 m/s, so the impact speed is 16.133938 m/s at 40.396333° below horizontal.

Result: The ball lands about 23.88 m away after 1.94 s, having peaked at 5.57 m. With the default 20 m/s at 45° from the ground, the range is 40.788649 m.

Understanding the result

On level ground, complementary angles such as 30° and 60° give the same range; the steeper one goes higher and stays up longer. Launching from a height shifts the best angle a little below 45°.

The impact speed is always at least the launch speed when landing at or below the launch level, because the drop adds energy. On level ground it equals the launch speed, and the impact angle mirrors the launch angle.

Assumptions and limitations

  • No air resistance, spin or wind.
  • Gravity is constant and points straight down (flat-Earth approximation), which is accurate for ranges far smaller than the planet.
  • The landing surface is flat and level, h₀ below the launch point.
  • Drag matters for fast, light or long-range projectiles such as golf balls, shuttlecocks and artillery; real ranges are shorter.
  • Landing on a slope or a raised target is not modeled; only a flat landing level at or below the launch point.
  • Very long ranges need Earth's curvature and changing gravity, which are ignored here.

Frequently asked questions

What launch angle gives the maximum range?

On level ground with no air resistance, 45°. Launching from above the landing level makes the best angle slightly lower, and air resistance in real life lowers it further.

Does the mass of the projectile matter?

Not without air resistance. Every object has the same acceleration due to gravity, so speed, angle and height alone decide the path.

Can I use a negative launch angle?

Yes, down to −90°, for something thrown downward from a height. A downward launch from ground level would land immediately, so the calculator asks for a launch height in that case.

How do I find horizontal projectile motion?

Set the angle to 0° and enter the launch height. The time of flight is then √(2h₀ ÷ g), and the range is the launch speed multiplied by that time.

How this calculator is checked

Method: Drag-free projectile kinematics with constant gravity: independent horizontal and vertical motion, landing time from the positive root of h₀ + v_y t − ½gt² = 0. The calculation is covered by automated tests with known reference values and edge cases. Last updated 2026-10-18. Found a problem? Tell us.

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