MotionLab

Projectile Motion Calculator

Launch speed and angle in — range, max height, and flight time out.

Launch speed (m/s)

Launch angle (°)

Gravity (m/s²)

Range (m)

40.775

R = v₀²·sin(2θ) / g

Max height (m): 10.194
Time of flight (s): 2.883

What you entered

Launch speed (m/s): 20Launch angle (°): 45Gravity (m/s²): 9.81

1
Horizontal velocity: vₓ = v₀·cos(θ)
20 × cos(45°)= 14.1421
2
Vertical velocity: vy = v₀·sin(θ)
20 × sin(45°)= 14.1421
3
Time of flight: t = 2vy ÷ g
2 × 14.1421 ÷ 9.81= 2.8832 s
4
Max height: h = vy² ÷ 2g
14.1421² ÷ (2 × 9.81)= 10.1937 m
5
Range: R = v₀²·sin(2θ) ÷ g
20² × sin(90°) ÷ 9.81= 40.7747 m

Result

Range (m): 40.775

Launched at 20 m/s and 45°, the projectile travels 40.77m, reaching a max height of 10.19m over 2.88s.

Projectile flight time, height and range by launch angle

Launched from ground level with no air resistance, at 9.81 m/s2. Range is horizontal distance to landing.

Launch speed (m/s)	Angle (deg)	Time of flight (s)	Max height (m)	Range (m)
10	15	0.528	0.341	5.097
20	30	2.039	5.097	35.312
20	45	2.883	10.194	40.775
20	60	3.531	15.291	35.312
30	45	4.325	22.936	91.743
50	45	7.208	63.710	254.842

The 30 and 60 degree rows at 20 m/s give exactly the same range of 35.312 m, which is the classic result: complementary angles produce identical range, with the steeper one simply spending longer in the air and going higher. Maximum range is always at 45 degrees, as the middle row shows. Doubling the launch speed from 25 to 50 would quadruple the range, since range depends on the square of speed. All of this ignores air resistance, which in reality lowers the optimum angle below 45 degrees and shortens the range considerably for light objects.

Splitting motion into two independent axes

Projectile motion is solved by treating horizontal and vertical motion completely separately — horizontal velocity stays constant (ignoring air resistance) while vertical velocity is governed entirely by gravity, and combining the two gives the full trajectory.

Why 45° maximizes range

For a given launch speed on flat ground, a 45° launch angle produces the maximum horizontal range — any steeper wastes energy going too high, any shallower doesn't stay airborne long enough, which is why this angle shows up constantly in sports and ballistics.

Frequently asked questions

A soccer ball is kicked at 20 m/s at a 30° angle. How far does it travel?

Range = v²sin(2θ)/g = 400×sin(60°)/9.81 ≈ 35.3 m. Max height = v²sin²(θ)/(2g) = 400×0.25/19.62 ≈ 5.1 m. Flight time ≈ 2.04 s.

Why does 45° give maximum range but not maximum height?

At 45°, the horizontal and vertical velocity components are balanced, maximizing range. A steeper angle (say 70°) sends the ball much higher but it doesn't travel as far horizontally. For maximum height, launch straight up (90°).

Does this calculator account for air resistance?

No — this uses the idealized equations (no drag). Real projectiles, especially light or fast ones, travel 20-50% shorter than the calculated range. For drag effects, see the drag force calculator.

A cannon on a cliff 50 m high fires horizontally at 100 m/s. Where does the shell land?

Fall time from 50 m: t = √(2h/g) = √(100/9.81) ≈ 3.19 s. Horizontal range = 100 × 3.19 = 319 m. The horizontal launch speed doesn't affect fall time — only height does. See the free fall calculator for drop problems.

How is projectile motion different from the speed-distance-time calculator?

Speed-distance-time handles straight-line motion at constant speed. Projectile motion involves two dimensions simultaneously — constant horizontal velocity plus vertical acceleration due to gravity — requiring trigonometry and separate x/y analysis.