Finding the Optimal Launch Angle

This is the classic optimization problem of intro physics: at what angle does a projectile go furthest? The answer is famous — but the derivation using calculus is gorgeous and the visualization makes the why obvious.

The Physics

From a level surface back to the same level, the range as a function of launch angle is:

$R(\theta) = \dfrac{v_0^{2}\,\sin(2\theta)}{g}$

For a fixed $v_0$ and $g$, the only thing that matters is $\sin(2\theta)$, which oscillates between 0 and 1. To find the maximum, we use calculus: set $\dfrac{dR}{d\theta} = 0$ and solve.

Step-by-Step Solution

Given:

• Launch speed: $v_0 = 15\;\text{m/s}$
• Gravity: $g = 9.81\;\text{m/s}^{2}$

Find: The launch angle $\theta_{\text{opt}}$ that maximizes the range $R$, and the value of $R$ at that angle.

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Step 1 — Write down the range function.

$R(\theta) = \dfrac{v_0^{2}\,\sin(2\theta)}{g} = \dfrac{(15)^{2}}{9.81}\,\sin(2\theta) = \dfrac{225}{9.81}\,\sin(2\theta) \approx 22.94\,\sin(2\theta)$

Step 2 — Differentiate with respect to $\theta$.

Apply the chain rule to $\sin(2\theta)$:

$\dfrac{dR}{d\theta} = \dfrac{v_0^{2}}{g} \cdot \dfrac{d}{d\theta}[\sin(2\theta)] = \dfrac{v_0^{2}}{g} \cdot 2\cos(2\theta) = \dfrac{2 v_0^{2}\cos(2\theta)}{g}$

Step 3 — Set the derivative to zero and solve.

$\dfrac{2 v_0^{2}\cos(2\theta)}{g} = 0$

The constants $v_0$ and $g$ aren't zero, so we need:

$\cos(2\theta) = 0$

$2\theta = 90° \quad\text{(taking the smallest positive solution)}$

$\boxed{\theta_{\text{opt}} = 45°}$

Step 4 — Confirm it's a maximum (second-derivative test).

$\dfrac{d^{2}R}{d\theta^{2}} = -\dfrac{4 v_0^{2}\sin(2\theta)}{g}$

At $\theta = 45°$, $\sin(2\theta) = \sin 90° = 1 > 0$, so the second derivative is negative — confirming a maximum.

Step 5 — Plug $\theta = 45°$ into the range equation.

$R_{\max} = \dfrac{v_0^{2}\sin(90°)}{g} = \dfrac{v_0^{2}}{g} = \dfrac{225}{9.81} \approx 22.94\;\text{m}$

We can also compute the corresponding peak height and flight time:

$H = \dfrac{v_0^{2}\sin^{2}(45°)}{2g} = \dfrac{225 \times 0.5}{19.62} \approx 5.74\;\text{m}$

$T = \dfrac{2 v_0\sin(45°)}{g} = \dfrac{2 \times 15 \times 0.7071}{9.81} \approx 2.162\;\text{s}$

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Answer: The optimal launch angle is $\theta_{\text{opt}} = 45°$, which produces the maximum horizontal range of $R_{\max} \approx 22.94\;\text{m}$. At this optimum, the ball reaches a peak height of about 5.74 m and flies for about 2.16 seconds.

Why 45°? An Intuition

You can think of it as a trade-off. Low angles give a lot of horizontal velocity but very little flight time. High angles give long flight times but tiny horizontal velocity. 45° is the perfect compromise — and the math agrees.

Note the symmetry: $R(\theta) = R(90° - \theta)$. So 30° and 60° give the same range, and so do 20° and 70°. The range curve $R(\theta)$ is symmetric about $\theta = 45°$.

Try It

• Sweep the angle slider — the range plot on the right shows $R$ vs $\theta$ live, with a yellow marker on the current angle.
• The peak of the curve is dead-center at 45°, marked by the green dashed line.
• Increase velocity — the entire curve scales up (since $R \propto v_0^{2}$) but the peak stays at 45°.