# Maximize time of flight and range at the same time in projectile motion

Given the equations of projectile motion (no air resistance), it is easy to find the launching angle theta that produces the maximum range. That angle is 45 degrees. The maximum time of flight is obtain instead for a launching angle theta=90 degrees: the projectile is launched straight up and the range is zero.

I would like to find the launch angle necessary to obtain the maximum range and maximum time of flight simultaneously. There must be a launch angle such that the obtained range and time of flight may not each be the maximum but together are the the largest. How would I set the problem up to find this special launch angle? Do I need express both time of flight and range as a function of theta,i.e. R(theta) and T(theta), multiply the two functions and set the derivative to zero? Should I find the maximum of the product R(theta)*T(theta) or the maximum of the sum R(theta)+T(theta)? Or something else?

• What you want to optimize for is a choice of yours. The problem with something like the sum of distance and time is that they have different units, which would make the result depending on the choice of units (and thus unphysical). The product doesn't suffer from that problem, but neither does the product of any arbitrary function of r with an arbitrary function of t. Commented Feb 26, 2016 at 20:43
• Example: say at 90° flight time is 2 seconds and horizontal distance is 0 meters vs. at 45° flight time/distance is 1.5 seconds and 10 meters vs. at 70° flight time/distance is 1.9 seconds and 6 meters. Is 1.5 seconds and 10 meters greater "maximum range and maximum time of flight" than 1.9 seconds and 6 meters? Only you can decide that. You can't simply add the two numbers together because if you measured distance in millimeters (for example) then distance would become way more important than time and you'd end up with 45° as your answer. Commented Feb 26, 2016 at 21:16
• Well, if you specify what you want to maximize, then it is math. Commented Feb 27, 2016 at 0:17
• Please answer which is "best": 1.5 seconds flight time and 10 meters flight distance or 1.9 seconds flight time and 6 meters flight distance. Commented Mar 2, 2016 at 18:40
• One example of such a function could be "f = c*distance + k*time" where c and k are the weighted constants with values you pick and units that make f dimensionless. Commented Mar 4, 2016 at 13:47

Tim is optimizing the sum of height and distance X of the parabolic trajectory, not the time of flight T = 2*usin(theta)/g so he does not answer the question. To obtain the same units you could instead optimize the sum of uT and X giving sin(theta) = sqrt(2/3) or theta = 60.8 degrees.

• I misread the question completely. Now my answer seems senseless! Commented Mar 5, 2016 at 14:54
• Actually the question itself is a little bit senseless since X only depends on u and T and not on theta.
– Jens
Commented Mar 5, 2016 at 15:02
• Since X = uTsqrt(1-(gT/2/u)^2) a good choice would be T = u*sqrt(2)/g which is the greatest T before X starts to decline.
– Jens
Commented Mar 5, 2016 at 15:30

If $v$ is the launch velocity and $\theta$ is the launch angle above the horizontal, the time of flight is $T=\frac {2v \sin \theta}g$ The distance of flight is $R=\frac {2v^2 \sin \theta \cos \theta}g$. It does not make sense to add them directly because the units do not match. We could define a constant $k$ with units distance/time and optimize $R+kT=\frac {2v}g(v \sin \theta \cos \theta + k \sin \theta)$ The constant $k$ says how important flight time is to us compared to range. If we set it to zero, we ignore the flight time, maximize the range, and get $\theta = 45^\circ$. If we make it very high, we essentially ignore the range, maximize the flight time, and get $\theta=90^\circ$. In between, we get a value in between. We could make $k$ dimensionless by defining $k'=\frac kv$, which is nice because $v$ sets the scale of the problem. Then we are maximizing $\sin \theta (1+k' \cos \theta)$, which has derivative $\cos \theta(1+k' \cos \theta)-k'\sin^2 \theta=\cos \theta+2k'\cos(2 \theta)-1$ and you can solve for $\theta$ as a function of $k'$

• what would k be if T and R were given equal importance? Commented Mar 11, 2016 at 19:17
• I don't think that question makes sense because $k$ is dimensional. The intuitive thing would be to take $k=1$ as giving them equal importance. If I measure T in seconds and R in mm, I get a very different answer than if I measure T in microseconds and R in km. If you set $k'=1$ you get $\theta = \arctan \frac 1{\sqrt 7}\approx 0.361367$ Commented Mar 11, 2016 at 20:12
• That is 20.7 degrees. I have to try and see what T and R are. Commented Mar 15, 2016 at 15:23
• The complementary of 20.7 degrees is 69.7 degrees and it should give the same range and longer time of flight... Commented Mar 15, 2016 at 15:24