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Friday, April 17, 2015

Maximizing The Trajectory Of A Projectile

Suppose we are asked to compute the launch angle which will maximize the arc-length of the trajectory for a projectile, assuming gravity is constant and is the only force on the projectile after the launch.

Eliminating the parameter t, we find the object's trajectory will be given by:

y=tan(θ)xgsec2(θ)2v20x2

dydx=tan(θ)gsec2(θ)v20x

s=v20gsin(2θ)01+(dydx)2dx

At this point I wanted to be slick and use the derivative form of the FTOC to differentiate directly, but I couldn't make it work. :lol:

So, here goes...

Let:

u=gv20sin(2θ)xdu=gv20sin(2θ)dx

and we have:

1+(tan(θ)gsec2(θ)v20x)2=1+(tan(θ)2tan(θ)u)2=cos2(θ)+sin2(θ)(12u)2cos2(θ)=

cos2(θ)+sin2(θ)(14u+4u2)cos2(θ)=cos2(θ)+sin2(θ)4usin2(θ)+4u2sin2(θ)cos2(θ)=

14usin2(θ)(1u)cos2(θ)

And now we may write:

s=1014usin2(θ)(1u)cos2(θ)v20sin(2θ)gdu

s=2v20sin(θ)g1014sin2(θ)u(1u)du

If we observe that this integral is symmetric about the line u=12, we may write:

s=4v20sin(θ)g12014sin2(θ)u(1u)du

Let:

v=4u(1u)dv=4(12u)du=41vdu and we have:

s=v20sin(θ)g101sin2(θ)v1vdv

Let:

w=1sin2(θ)v1vdw=(w2sin2(θ))22cos2(θ)wdv

and we have:

s=2v20sin(θ)cos2(θ)g1w2(w2sin2(θ))2dw

Let:

w=sin(θ)csc(α)dw=sin(θ)csc(α)cot(α)dα

and we have:

s=2v20cos2(θ)gθ0sec3(α)dα

With a bit of algebra, we may write this in a form we may integrate directly:

sec3(α)=12(sec3(α)+sec3(α))=12(sec3(α)+sec(α)(tan2(α)+1))=

12(sec3(α)+sec(α)tan2(α)+sec(α)tan(α)+sec(α)tan(α)+sec(α))=

12(sec(α)sec2(α)+sec(α)tan2(α)+sec(α)tan(α)+sec2(α)sec(α)+tan(α))=

12ddθ(sec(α)tan(α)+ln|sec(α)+tan(α)|)

and so we have (given 0<θ<π2):

s=v20cos2(θ)gθ0d(sec(α)tan(α)+ln(sec(α)+tan(α)))

s(θ)=v20cos2(θ)g(sec(θ)tan(θ)+ln(sec(θ)+tan(θ)))

s(θ)=v20g(sin(θ)+cos2(θ)ln(sec(θ)+tan(θ)))

Now, differentiating and equating to zero, we find:

s(θ)=v20g(cos(θ)+cos2(θ)sec(θ)tan(θ)+sec2(θ)sec(θ)+tan(θ)2cos(θ)sin(θ)ln(sec(θ)+tan(θ)))=0

s(θ)=2v20cos(θ)g(1sin(θ)ln(sec(θ)+tan(θ)))=0

And so we find we want to satisfy:

f(θ)=sin(θ)ln(sec(θ)+tan(θ))1=0

Using Newton's method, we will need:

f(θ)=tan(θ)+cos(θ)ln(sec(θ)+tan(θ))

And so Newton's method gives us the recursion:

θn+1=θnsin(θn)ln(sec(θn)+tan(θn))1tan(θn)+cos(θn)ln(sec(θn)+tan(θn))

Using θ0=π4 as the first guess, we find:

θ11.01751306067

θ20.986290147513

θ30.985515191499

θ40.985514737863

θ50.985514737862

θ60.985514737862

Converting from radians to degrees, we find the launch angle which maximizes the arc length is:

θ56.4658351275

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