Numford
Projectile motion

Range of a horizontal projectile

R = u √(2h/g). Thrown level at 10 m/s from 20 m with g = 10 m/s², the ball lands 20 m away after 2 s.

Projectile motion

Drag the launch arrow, or type the speed and angle. Slide the time to see the velocity at that moment. The dots are equal steps of time apart. Air resistance is ignored, as in NCERT.

Time to fall t2 s√(2h / g)
Range R20 mu √(2h / g)
Height h20 mthe drop
Velocity at 1 s14.14 m/svx = 10, vy = -10, 45° down
Pathy = 20 - 0.05x²y = h - gx² / 2u²
Hit speed22.36 m/s√(u² + 2gh)
Hit angle63.43°below the horizontal

A ball thrown level at 10 m/s from 20 m falls for 2 s and lands 20 m away at 22.36 m/s.

g (m/s²)
Show the formulas
Time to fall
t = √(2h / g) = √(2 × 20 / 10) = 2 s
Range
R = u t = 10 × 2 = 20 m
Hit speed
v = √(u² + 2gh) = √(10² + 2 × 10 × 20) = 22.36 m/s
Hit angle
tan β = gt / u = 20 / 10, so β = 63.43° below the horizontal
Velocity
vx = u cos θ = 10 m/s, vy = u sin θ - gt = -10 m/s at t = 1 s, so v = 14.14 m/s
Path
y = h - gx² / 2u², here y = 20 - 0.05x²
Working

Step by step

  1. Given u = 10 m/s, thrown level from h = 20 m, with g = 10 m/s².
  2. It falls like a dropped ball: h = ½gt², so t = √(2h / g) = √(2 × 20 / 10) = 2 s.
  3. Sideways it keeps u: R = u t = 10 × 2 = 20 m.
  4. On landing vx = 10 m/s and vy = gt = 10 × 2 = 20 m/s.
  5. Speed v = √(u² + 2gh) = √(10² + 20²) = 22.36 m/s (about 22.4 m/s), at tan⁻¹(20 / 10) = 63.43° below the horizontal.
Derivation

Why it works

  1. A ball thrown level from a height h starts with ux = u and uy = 0.
  2. It falls like a dropped ball: h = ½gt², so it takes t = √(2h/g) to reach the ground.
  3. Meanwhile it moves sideways at u the whole time, so the range is R = u √(2h/g).
  4. When it lands, vx = u and vy = gt = √(2gh), so the speed is √(u² + 2gh).
  5. It hits at tan β = vy / vx below the horizontal. A ball dropped at the same moment lands at the same time.
Questions

Horizontal projectile range formula: quick answers

What is the range formula for a horizontal projectile?

R = u √(2h/g), where u is the launch speed and h the height it is thrown from.

What is the range at 10 m/s from 20 m?

t = √(2 × 20 / 10) = 2 s, so R = 10 × 2 = 20 m.

How does the height change the range?

The range grows with √h. Four times the height doubles the time in the air and so doubles the range.

Try your own throw