Launch an idea
Projectile motion, explained.
Projectile motion separates into constant horizontal velocity and vertical acceleration. Gravity changes the vertical component of velocity; it does not change the horizontal component in this ideal model.
The relationship to remember
x = v₀ cos(θ)t · y = v₀ sin(θ)t − ½gt²
A worked example
At 20 m/s and 45° under 9.81 m/s² gravity, flight lasts about 2.88 s, the range is 40.77 m, and the peak is 10.19 m.
Inside the default experiment
- 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.
When this model applies
Point particle launched and landing at the same height; uniform gravity and no air resistance.
Make a prediction. Then test it.
Compare 30° and 60° at the same speed. What stays the same?
Try the interactive experiment ↗A little question. A clearer picture.
Frequently asked questions
Wondering about the why? Start here.
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.