at what height does horizontal launch (0 degree launch angle) begin to have greatest range?

Question

if ever?

Answer 1

Hello Nev!

Short Answer: Horizontal launch angles are only the optimal launch angles if your initial starting height is infinity (so, in reality, they are never optimal).

Longer Answer:

Here is a super generalized equation for the range (R) of a projectile based on its launch speed (v₀), its launch angle above horizontal (θ), the acceleration due to gravity (g), and the initial height (h).

R = [v₀cos(θ) / g] * [v₀sin(θ) + √(v₀²sin²(θ) + 2gh)]

You can prove this range expression yourself based on the projectile motion "displacement-velocity-acceleration" formulas if desired.

Now, if instead we wanted an expression for what launch angle (θ) would maximize the range for any given initial starting height (h), we'd need to do some calculus on the above expression. I'll omit that work unless you're interested in knowing the details and just jump straight to the below expression:

θ_{max range} = cos^-1 [ √(2gh + v₀²) / √(2gh + 2v₀²) ]

So if we want to analyze the conditions under which the optimal range launch angle is horizontal, then we want to consider what conditions cause θ_{max range} = 0°

0 = cos^-1 [ √(2gh + v₀²) / √(2gh + 2v₀²) ]

Take the cos() of both sides...

cos(0°) = √(2gh + v₀²) / √(2gh + 2v₀²)

1 = √(2gh + v₀²) / √(2gh + 2v₀²)

√(2gh + 2v₀²) = √(2gh + v₀²)

2gh + 2v₀² = 2gh + v₀²

2gh = 2gh - v₀²

So in order for a horizontal launch to have the maximum possible range, the two sides of that equation must be equal to each other -- which is clearly impossible (except in the trivial case of zero initial launch speed). However, this expression approaches being correct when 2gh >>> v₀² , so that the difference caused by the -v₀² term is inconsequential compared to the huge relative magnitude of the 2gh term.

To sum this all up, the only time that a horizontal launch will be the maximum range launch angle is when 2gh >>> v₀², or, put more clearly, in the limit as h --> infinity.

So, for all non-infinite starting heights, a non-horizontal launch angle will have a greater range than a horizontal launch angle.

Hope that helps!

--Zach