Projectile Motion Calculator
Range, maximum height, time of flight and trajectory plot — with step-by-step working.
Formula
Vertical: y(t) = h + v₀·sin(θ)·t − ½·g·t²
Time of flight (level ground, h = 0): T = 2·v₀·sin(θ) / g
Maximum height: H = h + (v₀·sin θ)² / (2g)
Range (level ground): R = v₀²·sin(2θ) / g
How to use
- Enter the initial launch speed in m/s, km/h, mph or ft/s.
- Enter the launch angle above the horizontal (degrees or radians).
- Enter the launch height above the landing surface. Use 0 for level-ground launches.
- Pick the gravity for your scenario — Earth, Moon, Mars or a custom value.
- Press Calculate. The trajectory is drawn and every output is given with units.
Physics behind projectile motion
A projectile, once released, is pulled only by gravity. The classical idealisation (no air resistance) splits the motion into two independent axes: horizontal, where velocity is constant, and vertical, where acceleration equals −g. This decomposition is the cornerstone of kinematics and is the reason the path is a parabola.
At launch, the velocity vector has magnitude v₀ and makes an angle θ with the horizontal. The components are vx = v₀·cos(θ) and vy = v₀·sin(θ). Because no horizontal force acts, vx stays constant throughout the flight; the projectile travels a horizontal distance x = vx·t. Vertically, gravity reduces vy linearly; at the top of the flight vy = 0. The height grows then shrinks, producing the familiar symmetric parabola for level-ground launches.
Range is maximised at 45° on level ground because sin(2θ) reaches its peak at θ = 45°. When the launch point is above the landing surface (h > 0), the optimal angle drops below 45° — the extra fall time gives the horizontal component more time to work.
Worked example
v₀ = 20 m/s, θ = 45°, h = 0, g = 9.80665 m/s²
vx = 20 cos 45° = 14.142 m/s vy = 20 sin 45° = 14.142 m/s T = 2·14.142 / 9.80665 ≈ 2.884 s H = 14.142² / (2·9.80665) ≈ 10.19 m R = 20²·sin(90°) / 9.80665 ≈ 40.77 m
Common mistakes
- Mixing degrees and radians. The formulas use the sine of the angle — make sure your calculator (and this tool's unit selector) agrees.
- Using g = 10 as a rounded value. For exam working 9.81 or 9.80665 gives more accurate answers; our tool defaults to CODATA 9.80665.
- Forgetting that R = v²·sin(2θ)/g is only valid when the launch height equals the landing height.
- Negative heights. The tool allows h = 0 or positive values; the projectile lands at y = 0.
Related tools
FAQs
How do I use projectile motion calculator range height time?
Punch in initial velocity v, launch angle θ, and (optionally) launch height. The calculator returns range, peak height, and total flight time using vacuum equations: R = v²sin(2θ)/g, H = v²sin²(θ)/(2g), T = 2v sin(θ)/g for level ground. So a ball launched at 20 m/s at 30° has R = 400 × sin(60°)/9.81 ≈ 35.3 m, H ≈ 5.1 m, T ≈ 2.04 s. These vacuum results overestimate real projectile motion noticeably for fast or light objects, but they're exact for short, dense projectiles with mild drag. Always check whether the question expects drag-free idealised motion.
How to calculate projectile range from velocity and angle?
For level launch, R = v²sin(2θ)/g, where v is initial speed, θ is launch angle, and g ≈ 9.81 m/s². Maximum range happens at θ = 45°, giving R_max = v²/g. So at 30 m/s, max range is 900/9.81 ≈ 91.7 m. At any angle θ, the range is reduced by sin(2θ). Note that 30° and 60° give the same range, since sin(60°) = sin(120°) — they're complementary trajectories. This is one of the cleanest results in elementary physics, but it assumes no air resistance and ground at the same height as launch.
How do I use time of flight projectile motion calculator?
For level launch and landing, T = 2v sin(θ)/g — twice the time to peak height. So a ball launched at 25 m/s at 40° stays in flight T = 50 × sin(40°)/9.81 ≈ 3.28 s. If launched from a height h with initial vertical velocity v_y = v sin(θ), use the kinematic equation 0 = h + v_y t − ½gt² and solve the quadratic for t. Calculators handle the algebra, but understanding where the quadratic comes from helps with edge cases. Total flight time controls horizontal distance, since x = v cos(θ) × t throughout the flight.
How do I use maximum height projectile formula?
Maximum height for level launch is H = v²sin²(θ)/(2g). At θ = 90° (straight up), this peaks at H = v²/(2g). At 45°, H = v²/(4g) — half of the straight-up value. Higher launch speeds and steeper angles both raise the peak. So a 30 m/s launch at 60° peaks at H = 900 × 0.75/19.62 ≈ 34.4 m. The formula comes from energy conservation applied to the vertical component, or equivalently from kinematics with v_y = 0 at the apex. Use it whenever you need to know if a projectile clears an obstacle.
How do I use horizontal range projectile motion formula?
Level-ground range comes out to R = v²sin(2θ)/g. The sin(2θ) factor peaks at θ = 45°, where 2θ = 90°. Speed enters squared, so doubling launch speed quadruples range. So 20 m/s at 45° gives R = 400/9.81 ≈ 40.8 m, while 40 m/s at 45° gives 163.1 m. The formula assumes launch and landing at the same height. If they differ, you'll need the full kinematic approach with vertical position equation. For projectiles launched up a hillside or off a cliff, the simple formula doesn't apply directly.
How do I use projectile motion with initial height calculator?
When the launch point is above the landing surface, the vertical position equation becomes y = h + v sin(θ) × t − ½ g t². Set y = 0 and solve the quadratic in t to find flight time, then horizontal distance is x = v cos(θ) × t. So a ball thrown horizontally (θ = 0) from a 5 m cliff at 10 m/s lands when 0 = 5 − 4.9t², giving t ≈ 1.01 s and range ≈ 10.1 m. The calculator runs the quadratic for you, but understanding the structure helps when checking whether your answer is reasonable.
How do I use launch angle for maximum range calculator?
On level ground in vacuum, the magic angle is exactly 45°, giving R_max = v²/g. The reason: R = v²sin(2θ)/g is maximised when sin(2θ) = 1, which happens at 2θ = 90°. If launch and landing heights differ, the optimum shifts: when launching from above, it drops below 45°; when launching from below up to a higher target, it goes above 45°. Air resistance also drops the optimum, sometimes well below 45°. The 45° rule is correct only in the idealised vacuum, level-ground case — useful as a starting point but rarely the full answer.