Motion · Topic 3 of 17

Projectile Motion

Tap any line to see why it matters

Physics · motion

Split a projectile into two independent motions: steady horizontal speed, and vertical free fall at 9.8 m/s². Time is the only thing they share.

01 · The methodHorizontal and vertical: solve separately

02 · HorizontalHorizontal: constant speed, x = vₓt

03 · VerticalVertical: free fall, 9.8 m/s² down

Projectile = launched object moving under gravity only, on a curved path must know

The path is a parabola (no air resistance)

Solve horizontal and vertical separately — time links them must know

Horizontal: no acceleration, constant vₓ, distance x = vₓt must know

Vertical: free fall, a = 9.8 m/s² down the whole flight must know

Time in the air comes from the vertical motion only — find it first

Launched sideways and dropped together → land together

At the peak: vertical velocity 0, horizontal velocity still vₓ

Max range at 45°; angles adding to 90° land equally far

Range = horizontal distance; faster launch → longer range

Angled launch: vₓ = v₀ cos θ, v₀ᵧ = v₀ sin θ

Horizontal launch from height h: t = √(2h ÷ g)

Off a 4.9 m ledge at 3 m/s: t = 1 s, lands 3 m out

6 m/s from 44.1 m: t = √9 = 3 s, lands 6 × 3 = 18 m out

20 m/s at 30°: vₓ ≈ 17.3, v₀ᵧ = 10 → aloft 2 × 1.02 ≈ 2.04 s, ≈ 35.3 m

Commonly confused

Distance uses vₓ = v₀ cos θ, not the full launch speed

Show 3 more facts

Range R = v₀² sin 2θ ÷ g (level ground)

Time of flight T = 2v₀ sin θ ÷ g (same launch and landing height)

With air resistance, best angle ≈ 38°, not 45°

Pearl

Vertical decides how long; horizontal decides how far.

Practice this topic, then test the whole section.

HESI and HESI A2 are registered trademarks of Elsevier. NurseSavvy is not affiliated with or endorsed by Elsevier.