Projectile Motion — Soccer Ball Kick (Worked Example)

Medium

Problem

A soccer ball is kicked from ground level with speed 20 m/s at 30° above the horizontal. Ignoring air resistance, find (a) the time of flight, (b) the maximum height, and (c) the horizontal range.

Trajectory of the kicked ball with computed range and height.

Given

v₀ = 20 m/s
θ = 30°
g = 9.81 m/s²
launch and landing at same height

Find

Time of flight t, max height h, range R.

Principle

Horizontal and vertical motions are independent: horizontal velocity stays constant while gravity accelerates the ball downward at g.

Formula

t = \frac{2v_0\sin\theta}{g}, \quad h = \frac{(v_0\sin\theta)^2}{2g}, \quad R = \frac{v_0^2\sin 2\theta}{g}

t = 2v₀sinθ/g; h = (v₀sinθ)²/2g; R = v₀²sin2θ/g

Solution

Vertical component: $v_{0y} = 20\sin30° = 20 \times 0.5 = 10\ \mathrm{m/s}$
Time of flight: $t = 2 \times 10 / 9.81 = 20/9.81 = 2.04\ \mathrm{s}$
Max height: $h = 10^2/(2 \times 9.81) = 100/19.62 = 5.10\ \mathrm{m}$
Range: $R = 20^2 \times \sin60° / 9.81 = 400 \times 0.8660 / 9.81 = 35.3\ \mathrm{m}$

t ≈ 2.04 s; h ≈ 5.10 m; R ≈ 35.3 m

Unit check

(m/s)²/(m/s²) = m for height and range; (m/s)/(m/s²) = s for time. All consistent.

What this means physically

The ball stays airborne about 2 seconds, climbs a bit over 5 m — higher than a basketball hoop — and lands 35 m away, roughly a third of a soccer field.

Common mistake

Using the full 20 m/s in the height formula instead of the vertical component (10 m/s). Only v₀sinθ fights gravity.