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Projectile Motion Calculator

Find the range, maximum height, time of flight and impact velocity of a projectile from its launch speed, angle and height. The trajectory is plotted to scale, and you can change gravity to see how the same throw would go on the Moon or Mars.

Physics · MotionUpdated Sources cited

0 to 90; negative values aim downwards

Result

Range
40.79 m
Maximum height
10.2 m
Time of flight
2.884 s
Impact speed
20 m/s
01020304005101520Horizontal distance (m)Height (m)max 10.2 m
Both axes use the same scale, so the curve shows the true shape of the path.
Horizontal velocity (constant)
14.14 m/s
Initial vertical velocity
14.14 m/s
Time to highest point
1.442 s
Impact angle (below horizontal)
45°
Trajectory data
Time (s)Distance (m)Height (m)
000
0.28844.0793.671
0.57688.1586.526
0.865312.248.566
1.15416.329.789
1.44220.3910.2
1.73124.479.789
2.01928.558.566
2.30732.636.526
2.59636.713.671
2.88440.790
Assumptions behind this result
  • No air resistance: the only force is gravity, acting straight down.
  • Gravity is uniform and the ground is flat; the projectile lands at height 0.
  • Angles are measured from the horizontal.

Step-by-step solution

  1. Resolve the launch velocity

    Split the launch velocity into horizontal and vertical components. With no air resistance, the horizontal component stays constant.

    vx=v0cos⁡θ=20cos⁡45∘=14.142 m/s,vy0=v0sin⁡θ=14.142 m/sv_x = v_0\cos\theta = 20\cos 45^\circ = 14.142\ \text{m/s},\quad v_{y0} = v_0\sin\theta = 14.142\ \text{m/s}

  2. Find the time of flight

    The vertical position is y = h₀ + v_{y0}t − ½gt². Setting y = 0 and taking the positive root of the quadratic gives the landing time.

    T=vy0+vy02+2gh0g=14.142+(14.142)2+2(9.8066)(0)9.8066=2.8842 sT = \frac{v_{y0} + \sqrt{v_{y0}^2 + 2gh_0}}{g} = \frac{14.142 + \sqrt{(14.142)^2 + 2(9.8066)(0)}}{9.8066} = 2.8842\ \text{s}

  3. Find the horizontal range

    The horizontal velocity is constant, so distance = velocity × time.

    R=vxT=(14.142)(2.8842)=40.789 mR = v_x T = (14.142)(2.8842) = 40.789\ \text{m}

  4. Find the maximum height

    At the highest point the vertical velocity is zero, which happens at t = v_{y0}/g.

    ttop=14.1429.8066=1.4421 s,H=h0+vy022g=0+(14.142)22(9.8066)=10.197 mt_{top} = \frac{14.142}{9.8066} = 1.4421\ \text{s},\quad H = h_0 + \frac{v_{y0}^2}{2g} = 0 + \frac{(14.142)^2}{2(9.8066)} = 10.197\ \text{m}

  5. Find the impact velocity

    At landing the horizontal velocity is unchanged and the vertical velocity is v_{y0} − gT.

    vy=14.142−(9.8066)(2.8842)=−14.142 m/s,∣v∣=vx2+vy2=20 m/sv_y = 14.142 - (9.8066)(2.8842) = -14.142\ \text{m/s},\quad |v| = \sqrt{v_x^2 + v_y^2} = 20\ \text{m/s}

What the calculator does

From the launch speed, angle and height, it calculates:

  • the range: the horizontal distance travelled before landing;
  • the maximum height above the ground;
  • the time of flight and the time to the highest point;
  • the impact speed and angle;
  • a trajectory graph, drawn to scale whenever the shape allows, with a table of points.

The formulas

The horizontal and vertical motions are independent. Horizontally there is no force, so the velocity is constant. Vertically the acceleration is −g.

vx=v0cos⁡θ,vy0=v0sin⁡θv_x = v_0\cos\theta, \qquad v_{y0} = v_0\sin\theta
x(t)=vxt,y(t)=h+vy0t−12gt2x(t) = v_x t, \qquad y(t) = h + v_{y0}t - \tfrac{1}{2}gt^2

Setting y = 0 and taking the positive root gives the time of flight:

T=vy0+vy02+2ghgT = \frac{v_{y0} + \sqrt{v_{y0}^2 + 2gh}}{g}

The range is R = vxT, and the highest point is reached at t = vy0/g, where

H=h+vy022gH = h + \frac{v_{y0}^2}{2g}

On level ground (h = 0) these reduce to the familiar R = v₀² sin 2θ / g and T = 2v₀ sin θ / g.

Worked example: 20 m/s at 45°

With g = 9.80665 m/s² and h = 0:

  1. Components: vx = vy0 = 20 cos 45° = 14.142 m/s.
  2. Time of flight: T = 2 × 14.142 / 9.80665 = 2.884 s.
  3. Range: R = 14.142 × 2.884 = 40.79 m.
  4. Maximum height: H = 14.142² / (2 × 9.80665) = 10.20 m.

At 30° the same throw travels 35.32 m, and at 60° it also travels 35.32 m, but it reaches 15.30 m high instead of 5.10 m.

Launching from a height

Throw a ball at 15 m/s and 30° from the top of a 10 m wall. It rises to 12.87 m above the ground, lands 30.98 m away after 2.38 s, and hits the ground at 20.5 m/s, 50.7° below the horizontal. It lands faster than it was thrown because it falls further than it rose.

Common mistakes

  • Using the level-ground range formula when the heights differ. R = v₀² sin 2θ / g only holds when launch and landing are at the same height.
  • Calculator in radians mode when using degrees.
  • Forgetting that horizontal velocity is constant. Only the vertical velocity changes.
  • Assuming the speed at the top is zero. Only the vertical component is zero there; the horizontal component remains.

Practical applications

  • Sport. Javelin, shot put, basketball and football kicks.
  • Engineering. Water jets, fountains and conveyor discharge.
  • Safety. Estimating where falling debris could land.
  • Teaching. Projectile motion is the classic example of separating motion into independent components.

Assumptions and limitations

The model ignores air resistance, spin, wind and the curvature and rotation of the Earth, and assumes the ground is flat and level with the landing point. Gravity is uniform, 9.80665 m/s² by default. The Moon, Mars and Jupiter presets use NASA’s mean surface values. Launch height is measured above the landing level, so landing above the launch point is not supported. For straight-line motion, use the kinematics calculator.

Frequently asked questions

What angle gives the maximum range?

45° when the projectile lands at the same height it was launched from and air resistance is ignored. From a raised launch point the best angle is a little less than 45°. With real air resistance it is lower still.

Why do 30° and 60° give the same range?

On level ground the range is v² sin 2θ / g, and sin 2θ is the same for complementary angles, since sin 60° = sin 120°. The 60° throw goes higher and stays in the air longer but covers the same distance.

Does the mass of the projectile matter?

Not without air resistance. Every object falls with the same acceleration g, so mass does not appear in any of the equations. In real air, heavier and more compact objects are slowed less.

How accurate is this for a real ball?

It ignores air resistance, so it overestimates the range, often by a lot for light or fast objects. It is a good model for dense, slow projectiles over short distances and is the standard model in physics courses.

Can I launch downwards?

Yes. Enter a negative angle together with a launch height, for example a ball thrown downwards from a cliff.

References

  1. Young, H. D. & Freedman, R. A. (2019). University Physics (15th ed.), chapter 3. Pearson.
  2. Halliday, D., Resnick, R. & Walker, J. (2018). Fundamentals of Physics (11th ed.), chapter 4. Wiley.
  3. NASA Goddard Space Flight Center. Planetary Fact Sheet (surface gravity values).

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