Numford
Projectile motion

Projectile thrown from a height

Thrown level at 20 m/s from 45 m with g = 10 m/s², the ball falls for 3 s, lands 60 m out and hits at 36.06 m/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 t3 s√(2h / g)
Range R60 mu √(2h / g)
Height h45 mthe drop
Velocity at 2 s28.28 m/svx = 20, vy = -20, 45° down
Pathy = 45 - 0.0125x²y = h - gx² / 2u²
Hit speed36.06 m/s√(u² + 2gh)
Hit angle56.31°below the horizontal

A ball thrown level at 20 m/s from 45 m falls for 3 s and lands 60 m away at 36.06 m/s.

g (m/s²)
Show the formulas
Time to fall
t = √(2h / g) = √(2 × 45 / 10) = 3 s
Range
R = u t = 20 × 3 = 60 m
Hit speed
v = √(u² + 2gh) = √(20² + 2 × 10 × 45) = 36.06 m/s
Hit angle
tan β = gt / u = 30 / 20, so β = 56.31° below the horizontal
Velocity
vx = u cos θ = 20 m/s, vy = u sin θ - gt = -20 m/s at t = 2 s, so v = 28.28 m/s
Path
y = h - gx² / 2u², here y = 45 - 0.0125x²
Working

Step by step

  1. Given u = 20 m/s, thrown level from h = 45 m, with g = 10 m/s².
  2. It falls like a dropped ball: h = ½gt², so t = √(2h / g) = √(2 × 45 / 10) = 3 s.
  3. Sideways it keeps u: R = u t = 20 × 3 = 60 m.
  4. On landing vx = 20 m/s and vy = gt = 10 × 3 = 30 m/s.
  5. Speed v = √(u² + 2gh) = √(20² + 30²) = 36.06 m/s (about 36.1 m/s), at tan⁻¹(30 / 20) = 56.31° 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

Projectile from a height: quick answers

How do you solve a projectile thrown from a height?

Treat the two directions apart. Use the vertical fall to find the time, then multiply the horizontal speed by that time for the range.

What is the speed on landing at 20 m/s from 45 m?

v = √(u² + 2gh) = √(20² + 2 × 10 × 45) = 36.06 m/s, at 56.31° below the horizontal.

Does the landing speed depend on the throw angle?

Its size does not: by energy, v = √(u² + 2gh) for any angle. Its direction and the time in the air do.

Try your own throw