Projectile motion: optimal launch angle from a platform

In the absence of air resistance, a launch angle of 45° maximises range. If the projectile is instead launched from a raised platform, the optimal launch angle still has a closed-form expression, although more algebra is required to compute it.

Solution

Suppose the projectile is launched at speed $u$ and angle $\phi$ from height $h$ above the ground, which has gravitational field strength $g$, so that the motion is given by

The flight time is the positive solution to the quadratic $y (t) = 0$, which is

where $C = 2 g h / u^2$ is the dimensionless ratio between the initial potential and kinetic energies of the projectile. Substituting the flight time into $x (t)$ gives the range

To maximise the range $R$ with respect to the launch angle $\phi$, we compute the derivative $\pd R / {\pd\phi}$. After some algebra, we obtain

where for brevity $\beta = \sin\phi$. We then carefully consider the ways in which $\pd R / {\pd\phi}$ can vanish:

1. If $C = \infty$, then

   $$\frac{\pd R}{\pd\phi} = \frac{2h}{C} \colb{\roundbr{\sqrt{C}}} \colr{\squarebr{-\beta}} = - 2\beta \cdot \frac{h}{\sqrt{C}},$$

   which vanishes assuming $h$ is finite. But since $C = 2 g h / u^2$ is infinite, this only occurs if $g = \infty$ (infinitely strong gravity) or $u = 0$ (zero launch speed), and in either case the range is zero, i.e. $R$ is minimised.

2. If $\colb C = 0$ and $\colb \beta < 0$, then

   $$\begin{aligned} \frac{\pd R}{\pd\phi} &= \frac{2h}{C} \colb{\roundbr{ \beta + (-\beta) \roundbr{1 + \frac{C}{2 \beta^2}} }} \colr{\squarebr{ \frac{1 - \beta^2}{-\beta} + \frac{\beta^2}{-\beta} }} \\[\tallspace] &= \frac{2h}{C} \colb{\roundbr{ - \frac{C}{2 \beta} }} \colr{\squarebr{ - \frac{1}{\beta} }} \\[\tallspace] &= \frac{h}{\beta^2}, \end{aligned}$$

   which vanishes only if $h = 0$. But since $\phi = \sin^{-1}\beta < 0$, this corresponds to launching the projectile downwards starting from ground level, and again the range is zero.

3. If $\colr{ \squarebr{\frac{1 - \beta^2}{\sqrt{\beta^2 + C}} - \beta} = 0 }$, then $\pd R / {\pd\phi}$ vanishes unconditionally, and we have

   $$\begin{aligned} (1 - \beta^2)^2 &= \beta^2 (\beta^2 + C) \\ \beta &= \frac{1}{\sqrt{C + 2}}, \end{aligned}$$

   which indeed corresponds to the positive maximum range, which is

   $$\begin{aligned} R &= \frac{u^2}{g} \sqrt{1 - \beta^2} \roundbr{\beta + \sqrt{\beta^2 + C}} \\[\tallspace] &= \frac{u^2}{g} \sqrt{C + 1}. \end{aligned}$$

Result

Hence the optimal launch angle for a projectile launched at speed $u$ from height $h$ in a gravitational field of strength $g$ is

achieving a maximum range of

In particular:

• For $h = 0$, we recover $\phi = \sin^{-1} (1/\sqrt{2}) = 45\degree$ and $R = u^2 / g$, which are the optimal angle and maximum range respectively for a projectile launched from ground level. For small $h$ we have
  $$\phi \asy 45\degree - \frac{g h}{2 u^2} \cdot \frac{180\degree}{\pi}$$
  and
  $$R \asy \frac{u^2}{g} + h.$$
• For $h = \infty$, the optimal angle is $\phi = \sin^{-1} (1 / \sqrt{\infty}) = 0$ and the maximum range is $R = u^2 / g \cdot \sqrt{\infty} = \infty$. Projectiles should be launched almost horizontally from very tall platforms; asymptotically we have
  $$\phi \asy \frac{1}{\sqrt{2 g h / u^2}} = \frac{u}{\sqrt{2 g h}} \cdot \frac{180 \degree}{\pi}$$
  and
  $$R \asy \frac{u^2}{g} \sqrt{2 g h / u^2} = u \sqrt{\frac{2 h}{g}}.$$

Finally, note that the optimal angle depends only on the dimensionless ratio $C = 2 g h / u^2$. In fact this may be shown using dimensional analysis without actually having to solve the problem; the only parameters are $u$, $g$, and $h$, so the only dimensionless group (up to a power) is $g h / u^2$.

See also

• Projectile motion: optimal launch angle for weak quadratic drag

Cite this page

Conway (2022). Projectile motion: optimal launch angle from a platform. <https://yawnoc.github.io/math/projectile-platform> Accessed 2025-06-04.