Projectile motion

Horizontal projectile motion

A stone thrown level at 15 m/s from a 490 m cliff, with g = 9.8 m/s², lands after 10 s, 150 m out, at 99.14 m/s (about 99.1 m/s) and 81.3° below the horizontal.

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 5 s

Time to fall t10 s√(2h / g)

Range R150 mu √(2h / g)

Height h490 mthe drop

Velocity at 5 s51.24 m/svx = 15, vy = -49, 72.98° down

Pathy = 490 - 0.0218x²y = h - gx² / 2u²

Hit speed99.14 m/s√(u² + 2gh)

Hit angle81.3°below the horizontal

A ball thrown level at 15 m/s from 490 m falls for 10 s and lands 150 m away at 99.14 m/s.

Show the formulas

Time to fall: t = √(2h / g) = √(2 × 490 / 9.8) = 10 s
Range: R = u t = 15 × 10 = 150 m
Hit speed: v = √(u² + 2gh) = √(15² + 2 × 9.8 × 490) = 99.14 m/s
Hit angle: tan β = gt / u = 98 / 15, so β = 81.3° below the horizontal
Velocity: vx = u cos θ = 15 m/s, vy = u sin θ - gt = -49 m/s at t = 5 s, so v = 51.24 m/s
Path: y = h - gx² / 2u², here y = 490 - 0.0218x²

Working

Step by step

Given u = 15 m/s, thrown level from h = 490 m, with g = 9.8 m/s².
It falls like a dropped ball: h = ½gt², so t = √(2h / g) = √(2 × 490 / 9.8) = 10 s.
Sideways it keeps u: R = u t = 15 × 10 = 150 m.
On landing vx = 15 m/s and vy = gt = 9.8 × 10 = 98 m/s.
Speed v = √(u² + 2gh) = √(15² + 98²) = 99.14 m/s (about 99.1 m/s), at tan⁻¹(98 / 15) = 81.3° below the horizontal.

Derivation

Why it works

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

Questions