Projectile motion: how far and high?
Type a launch speed, a launch angle and, if it starts above the ground, a launch height. The projectile motion calculator gives how far it lands, how long it flies, how high it goes and how fast it hits, and draws its path.
- Range
- 133.821 ft
It lands 133.821 ft away after 2.88419 s, reaching 33.4553 ft at the top.
- Time of flight (s)
- 2.88419
- Maximum height
- 33.4553 ft
- Time to the top (s)
- 1.4421
- Impact speed
- 65.6168 ft/s
- Impact angle (° below horizontal)
- 45
Range: 133.821 ft. It lands 133.821 ft away after 2.88419 s, reaching 33.4553 ft at the top.
The path of the projectile (meters)
How to calculate
Works out a projectile’s range, time of flight, maximum height and impact speed from its launch speed, angle and height, with no air resistance, and draws its path.
Example with the default inputs (Launch speed 65.6167979002625 ft/s, Launch angle 45 °, Launch height 0 ft, Gravity (m/s²) 9.80665): It lands 133.821 ft away after 2.88419 s, reaching 33.4553 ft at the top.
Method: vₓ = v₀ cos θ, vᵧ = v₀ sin θ; t = (vᵧ + √(vᵧ² + 2gh₀)) ÷ g; R = vₓ t; h_max = h₀ + vᵧ² ÷ 2g; v_impact = √(v₀² + 2gh₀).
- No air resistance, a flat landing level, and the same gravity everywhere along the path.
- The launch height is above the landing level; the launch angle is 0° to 90° above the horizontal.
Worked examples
Each example is checked against the calculator on every build.
- Launch speed 229.7 ft/s, Launch angle 75°, Launch height 0 ft, Gravity (m/s²) 9.8 gives Maximum height 765.266 ft, Time to the top (s) 6.89947, Range 820.21 ft.Source: OpenStax, University Physics Volume 1, §4.3 Projectile Motion (x = v₀ₓt, y = v₀ᵧt − ½gt², h = v₀ᵧ² ÷ 2g, R = v₀² sin 2θ₀ ÷ g), https://openstax.org/books/university-physics-volume-1/pages/4-3-projectile-motion (Example 4.7: h = 233 m, t = 6.90 s to the top)
- Launch speed 65.62 ft/s, Launch angle 45°, Launch height 0 ft, Gravity (m/s²) 9.80665 gives Range 133.821 ft, Time of flight (s) 2.884193, Maximum height 33.4553 ft.Source: OpenStax, University Physics Volume 1, §4.3 Projectile Motion (x = v₀ₓt, y = v₀ᵧt − ½gt², h = v₀ᵧ² ÷ 2g, R = v₀² sin 2θ₀ ÷ g), https://openstax.org/books/university-physics-volume-1/pages/4-3-projectile-motion
- Launch speed 32.81 ft/s, Launch angle 0°, Launch height 65.62 ft, Gravity (m/s²) 9.8 gives Time of flight (s) 2.020305, Range 66.283 ft, Impact speed 72.7725 ft/s.Source: OpenStax, University Physics Volume 1, §4.3 Projectile Motion (x = v₀ₓt, y = v₀ᵧt − ½gt², h = v₀ᵧ² ÷ 2g, R = v₀² sin 2θ₀ ÷ g), https://openstax.org/books/university-physics-volume-1/pages/4-3-projectile-motion
- Launch speed 98.43 ft/s, Launch angle 90°, Launch height 0 ft, Gravity (m/s²) 9.80665 gives Range 0 ft, Time of flight (s) 6.118297, Impact angle (° below horizontal) 90.Source: OpenStax, University Physics Volume 1, §4.3 Projectile Motion (x = v₀ₓt, y = v₀ᵧt − ½gt², h = v₀ᵧ² ÷ 2g, R = v₀² sin 2θ₀ ÷ g), https://openstax.org/books/university-physics-volume-1/pages/4-3-projectile-motion
How it works
Launch speed v₀ (m/s), launch angle θ above the horizontal, launch height h₀ (m) above the landing level, gravity g (m/s²):
- Speed parts: vₓ = v₀ cos θ (sideways), vᵧ = v₀ sin θ (upward). At exactly 90°, cos θ is taken as 0 and sin θ as 1; at 0°, sin θ is 0.
- Time of flight: t = (vᵧ + √(vᵧ² + 2 g h₀)) ÷ g, the time to come back down to the landing level.
- Range: R = vₓ × t. On level ground this is v₀² sin 2θ ÷ g.
- Maximum height: h_max = h₀ + vᵧ² ÷ (2g), reached at t = vᵧ ÷ g.
- Impact speed: √(v₀² + 2 g h₀); impact angle below the horizontal: atan2(√(vᵧ² + 2 g h₀), vₓ), in degrees.
The chart draws the path y = h₀ + x tan θ − g x² ÷ (2 vₓ²) from the launch to the landing point, in meters.
Units: 1 km/h = 1/3.6 m/s; 1 mph = 0.44704 m/s; 1 ft/s = 0.3048 m/s; 1° = π/180 rad; 1 ft = 0.3048 m; 1 in = 0.0254 m; 1 cm = 0.01 m.
Output format. Every value shows 6 significant figures. The range and maximum height show in feet and the impact speed in ft/s (meters and m/s in Metric); times are in seconds.
When there is no answer. An angle above 0 so small that the flight time or range is below the smallest number a computer can hold.
Assumptions
- No air resistance, a flat landing level, and the same gravity all along the path; the projectile is a point.
- Launch speed 0.001 to 100,000 m/s; angle 0° to 90°; launch height 0 to 100,000 m; gravity 0.001 to 1,000 m/s².
Worked examples by hand
A firework at 70.0 m/s and 75.0°, g = 9.8 (OpenStax Example 4.7). vᵧ = 70 sin 75° = 67.6148 m/s. Time to the top 67.6148 ÷ 9.8 = 6.89947 s; maximum height 67.6148² ÷ 19.6 = 233.253 m; level-ground range 70² × sin 150° ÷ 9.8 = 4,900 × 0.5 ÷ 9.8 = 250 m.
20 m/s at 45° on Earth (the default). R = 400 × sin 90° ÷ 9.80665 = 40.7886 m; t = 2 × 20 × sin 45° ÷ 9.80665 = 2.88419 s; h_max = (14.1421)² ÷ 19.6133 = 10.1972 m.
Thrown flat at 10 m/s from 20 m up, g = 9.8. vᵧ = 0, so t = √(2 × 9.8 × 20) ÷ 9.8 = √(2 × 20 ÷ 9.8) = 2.02031 s; R = 10 × 2.02031 = 20.2031 m; impact speed √(100 + 392) = 22.1811 m/s.
Straight up at 30 m/s. t = 2 × 30 ÷ 9.80665 = 6.1183 s; the range is 0 and it comes down at 90°.
Other questions people ask
What is the range formula for projectile motion?
On level ground, R = v₀² × sin 2θ ÷ g. A projectile at 20 m/s and 45° on Earth lands 20² × sin 90° ÷ 9.80665 = 40.79 m away. From a height the page uses the full flight time instead, because the level-ground formula no longer holds.
What angle gives the longest range?
45° on level ground, with no air resistance, because sin 2θ is largest at θ = 45°. From a raised launch the best angle is a little less than 45°. Angles that add to 90°, such as 30° and 60°, give the same level-ground range.
How long is a projectile in the air?
t = (v₀ sin θ + √((v₀ sin θ)² + 2 g h₀)) ÷ g, where h₀ is the launch height. On level ground this is 2 v₀ sin θ ÷ g.
How high does a projectile go?
The top of the path is (v₀ sin θ)² ÷ 2g above the launch point. A firework shell launched at 70 m/s and 75° (g = 9.8 m/s²) climbs 233 m in 6.90 s (OpenStax Example 4.7).
Does mass change projectile motion?
Not without air resistance: every mass falls with the same acceleration g, so the path depends only on the speed, angle, height and gravity.
Why is the real range shorter?
Air resistance slows the projectile, so a real ball or shell lands short of the ideal range, often by a lot at high speeds. The page leaves air resistance out.