Projectile motion simulator
Launch a ball at any speed and angle and this projectile motion simulator calculates its range, maximum height and time of flight step by step. For a launch and landing at the same height with no air resistance, range R = v₀² sin 2θ / g, maximum height H = v₀² sin²θ / (2g) and time of flight T = 2v₀ sin θ / g.
What if you changed…
Compare 30° and 60° at the same speed. What stays the same?
See the calculation
- Horizontal velocity = 20 cos(45°) = 14.14 m/s.
- Vertical launch velocity = 20 sin(45°) = 14.14 m/s.
- Flight time = 2vᵧ/g = 2.88 s. Range = vₓ × time = 40.77 m.
What this model assumes
Point particle launched and landing at the same height; uniform gravity and no air resistance.
Keep an eye on this
A 45° launch gives maximum range only with a level landing and no air resistance.
Understand the physics ↗Formulas used in this simulation
- Horizontal velocity
- vₓ = v₀ cos θ
- Constant for the whole flight
- Initial vertical velocity
- vᵧ = v₀ sin θ
- Gravity reduces it by g each second
- Position at time t
- x = v₀ cos θ · t, y = v₀ sin θ · t − ½gt²
- The path is a parabola
- Time of flight
- T = 2v₀ sin θ / g
- Level launch and landing
- Maximum height
- H = v₀² sin²θ / (2g)
- Reached halfway through the flight
- Range
- R = v₀² sin 2θ / g
- Largest at θ = 45°
- Horizontal launch from height h
- t = √(2h/g), x = v₀√(2h/g)
- Launched horizontally, θ = 0
Projectile motion: common questions
Wondering about the why? Start here.
How do you calculate the range of a projectile?
For a level launch with no air resistance, multiply the horizontal velocity by the flight time: R = vₓT = v₀² sin 2θ / g. At 20 m/s and 45° on Earth, R = 400 / 9.81 ≈ 40.8 m.
What is the formula for the maximum height of a projectile?
H = v₀² sin²θ / (2g). Only the vertical part of the launch velocity matters, because the horizontal part does not work against gravity.
How do I solve a horizontal projectile problem?
Treat the two directions separately. The fall time depends only on the height, t = √(2h/g), and the horizontal distance is x = v₀t. A ball rolled off a 5 m ledge at 3 m/s falls for about 1.01 s and lands about 3.03 m out.
Why is 45° the best launch angle here?
For a level landing, range is v₀² sin(2θ)/g. The sine term reaches 1 when 2θ is 90°, so θ is 45°.
Does a heavier ball travel farther?
Mass does not appear in this ideal model. Air resistance can make real trajectories depend on mass, size, and shape.