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 yyyy-mm-dd.