Optimal Launch Angle

It is a truth universally acknowledged that the optimal angle of launch of a projectile is 45 degrees to the horizontal. That is, when launched at this angle, the projectile will travel the greatest transverse distance before hitting the ground.cliff

Then if you were standing on a cliff, and threw a stone into the sea at this ‘optimal angle’, would it hit the water at the greatest possible distance from the base of the cliff?

The answer is no.

One of the several assumptions made when deriving the quoted result is that the projectile is launched from the height at which it lands. This premise is palpably false when we hurl something off a cliff. It is this tacit constraint that we’re going to loosen in this post.

The question is: what is the optimal launch angle when the projectile is launched from a distance y_0 above the ground?


We’ll state without proof the expressions for the coordinates of the stone as a function of time:



The nomenclature: v_0 is the speed of launch, \theta is the launch angle defined with respect to the x-axis (the horizontal), g is the magnitude of acceleration due to gravity and t is the time elapsed since launch. The assumptions: the stone suffers no dissipative forces and the gravitational field of earth is uniform in space and time over the range of the stone’s motion.

We first need to work out when the stone lands in the sea. This time is labelled \tau, and satisfies

y(\tau)=0,\ \tau>0

Then to find \tau it’s necessary to solve the quadratic equation


The positive solution is

\displaystyle \tau=\frac{v_0\sin\theta+\sqrt{v_0^2\sin^2\theta+2gy_0}}{g}

So the range R of the stone, the transverse distance at which it strikes the sea, is


\displaystyle R=v_0\cos\theta\Bigg(\frac{v_0\sin\theta+\sqrt{v_0^2\sin^2\theta+2gy_0}}{g}\Bigg)

or, after pulling the factor of v_0 out of the root,

\displaystyle R(\theta)=\frac{v_0^2}{g}\big(\sin\theta+\sqrt{\sin^2\theta+2\lambda}\big)\cos\theta




To find the launch angle that maximises this function, we set its derivative equal to 0

\displaystyle \frac{dR}{d\theta}=0

This will get a little messy, but tidies up quickly thereafter:

\displaystyle \frac{dR}{d\theta}=\Bigg(\cos\theta+\frac{\sin\theta\cos\theta}{\sqrt{\sin^2\theta+2\lambda}}\Bigg)\cos\theta-\big(\sin\theta+\sqrt{\sin^2\theta+2\lambda}\big)\sin\theta=0

\displaystyle \Bigg(1+\frac{\sin\theta}{\sqrt{\sin^2\theta+2\lambda}}\Bigg)\cos^2\theta-\big(\sin\theta+\sqrt{\sin^2\theta+2\lambda}\big)\sin\theta=0

Multiplying through by the square root gives

\displaystyle \big(\sqrt{\sin^2\theta+2\lambda}+\sin\theta\big)\cos^2\theta-\sqrt{\sin^2\theta+2\lambda}\big(\sin\theta+\sqrt{\sin^2\theta+2\lambda}\big)\sin\theta



For this equation to be satisfied, either bracket can be equal to 0.

For the second bracket to be equal to 0, we can see that \lambda must be 0. From the definition of \lambda, this means either gravity must be switched off, the stone is launched beyond light speed or the cliff’s height is 0 (precisely the instance we’re avoiding). None of these situations is of interest to us, so instead we set the first bracket equal to 0:


We rearrange and square to kill the square root:


We then replace the cosine squared term:


For a second, it looks like we’re going to have to solve a quartic, but happily the highest powers cancel:



\displaystyle \sin\theta=\frac{1}{\sqrt{2(1+\lambda)}}

Hence the optimal launch angle is

\displaystyle \theta=\arcsin\Bigg[\sqrt{2\Big(1+\frac{gy_0}{v_0^2}\Big)}\Bigg]^{-\frac{1}{2}}


How does this vary as a function of the launch height y_0?

If y_0 is equal to 0, the optimal launch angle is

\displaystyle \theta=\arcsin\Big(\frac{1}{\sqrt{2}}\Big)=45^\circ

So we can recover the traditional result. But as y_0 increases, the argument of the arcsine function decreases: as the launch height increases from 0, the optimal launch angle decreases from 45 degrees.


Can we use negative values of y_0? The answer is yes, provided



\displaystyle y_0 > \frac{v_0^2}{2g}

If the launch height does not satisfy this inequality, the arcsine returns an imaginary value. Why? Because the stone was not given enough initial energy to reach the height y=0 at all, so there does not exist a real optimal value of \theta. If y_0<0, the optimal launch angle \theta increases instead.


To give an idea of the magnitude of the effect, the optimal launch angle of a javelin launched at 30 metres per second from a cliff of height 100 metres is around 30 degrees to the horizontal. Try some examples for yourself to get an idea of how much the real answer deviates from the traditional one.


Here’s the real question: why does the launch angle decrease for increasing y_0? Let’s do away with maths (hooray!) and look for a physical explanation.

It’s helpful to ask: why is the ‘traditional’ optimal launch angle 45 degrees? Why not higher? Or lower?

Launching the projectile at a low angle means it covers ground quickly, but doesn’t spend long travelling before it lands. Launching the projectile at a high angle means it travels for a long time, but doesn’t cover ground particularly quickly. Launching halfway between the two is a compromise: the projectile covers ground fairly quickly for a fairly long time, and these two effects combine to give what is actually the greatest transverse range.

The diagram below shows the trajectory of a projectile launched at 30, 45 and 60 degrees. The one launched at 30 degrees covers ground fastest, but lands quickly. The one launched at 60 degrees spends longest in the air, but has a small horizontal component of velocity. The one launched at 45 degrees is a trade-off, and travels furthest.pedagogy

So what’s the difference if you launch from above the ground?

If you launch from very high up, the projectile you throw is bound to spend longer in the air than if it had been launched from the ground. Therefore it is less important to give the projectile its initial kick in the vertical direction, because the extra air-time you give it in doing so will pale in comparison to the air-time it has by virtue of being launched from very high up. It is more efficient to put the greater component of velocity in the horizontal direction so the projectile covers ground more quickly.

Can you think of an explanation for the other case? Why does \theta increase from 45^\circ if y_0<0?


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