Angled Projectile Launch — Soccer Ball at 45°

When you launch a ball at an angle $\theta$ above the horizontal with speed $v_0$, the velocity splits into two components — and each one obeys an independent 1D motion. This problem shows you how to compute the three big numbers of any projectile launch: time of flight, maximum height, and range.

The Physics

Decompose the launch velocity into horizontal and vertical components:

$v_{0x} = v_0 \cos\theta \qquad v_{0y} = v_0 \sin\theta$

The horizontal component never changes. The vertical component decelerates under gravity, hits zero at the peak, then accelerates back down. This gives three closed-form formulas:

$T = \dfrac{2 v_0 \sin\theta}{g} \qquad H = \dfrac{v_0^{2}\sin^{2}\theta}{2g} \qquad R = \dfrac{v_0^{2}\sin(2\theta)}{g}$

The range formula uses the identity $2\sin\theta\cos\theta = \sin(2\theta)$, which is why 45° gives the maximum range — $\sin(2\theta)$ peaks when $2\theta = 90°$.

Step-by-Step Solution

Given:

• Initial speed: $v_0 = 30\;\text{m/s}$
• Launch angle: $\theta = 45°$
• Gravity: $g = 9.81\;\text{m/s}^{2}$

Find: Maximum height $H$ and horizontal range $R$.

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Step 1 — Decompose the velocity into components.

$v_{0x} = v_0\cos\theta = 30\cos 45° = 30 \times \dfrac{\sqrt{2}}{2} \approx 21.21\;\text{m/s}$

$v_{0y} = v_0\sin\theta = 30\sin 45° = 30 \times \dfrac{\sqrt{2}}{2} \approx 21.21\;\text{m/s}$

(At 45° they're equal — that's the special property of this angle.)

Step 2 — Find the time to peak.

The ball reaches its peak when $v_y = 0$. Starting from $v_y(t) = v_{0y} - g\,t$:

$v_{0y} - g\,t_{\text{up}} = 0$

$t_{\text{up}} = \dfrac{v_{0y}}{g} = \dfrac{21.21}{9.81} \approx 2.163\;\text{s}$

Step 3 — Find the total flight time.

By symmetry, going up and coming down take the same time:

$T = 2\,t_{\text{up}} = 2 \times 2.163 \approx 4.325\;\text{s}$

Step 4 — Find the maximum height.

Plug $t_{\text{up}}$ into the vertical position equation $y(t) = v_{0y}\,t - \tfrac{1}{2}\,g\,t^{2}$:

$H = v_{0y}\,t_{\text{up}} - \tfrac{1}{2}\,g\,t_{\text{up}}^{2}$

$H = (21.21)(2.163) - \tfrac{1}{2}(9.81)(2.163)^{2}$

$H = 45.88 - 22.94 \approx 22.94\;\text{m}$

(Or directly with the formula: $H = v_{0y}^{2}/(2g) = 449.9/19.62 \approx 22.94\;\text{m}$.)

Step 5 — Find the horizontal range.

The horizontal velocity is constant at $v_{0x} = 21.21\;\text{m/s}$, and the ball is in the air for $T = 4.325\;\text{s}$:

$R = v_{0x}\,T = 21.21 \times 4.325 \approx 91.74\;\text{m}$

(Or directly with the formula: $R = v_0^{2}\sin(2\theta)/g = 900\sin 90°/9.81 = 900/9.81 \approx 91.74\;\text{m}$.)

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Answer: The soccer ball reaches a maximum height of ≈ 22.94 m and a horizontal range of ≈ 91.74 m, with a total flight time of ≈ 4.33 s.

Try It

• Sweep the angle slider — notice the range is symmetric about 45°. For example, 30° and 60° give the same range (about 79.5 m at 30 m/s).
• Crank up velocity — range scales with $v_0^{2}$, so doubling the speed quadruples the range.
• Watch the peak marker rise and fall — at 45° you trade height for range; at 80° you go very high but barely move.