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Calculus/calculating maximum length of arc on projectiles

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Expert: Alon Mandes - 12/5/2008


Question
Dear Sir :
Am jammed with this , I have been asked to calculate the maximum length of the arc of a projectile (general formula) using integration , I know i have to start with parabla equation i think but can not get any where , can you help please.

Answer
Ok, our projectile has the form r(t)=[Voxt,Voyt-(g/2)t^2].
(" g is 9.8m/sec^2 "). Thus, r'(t)=[Vox,Voy-gt]. The arc length
is defined as S=∫√[x'(t)^2+y'(t)^2]dt..Where t goes from 0 to T,
the whole period of the projectile. Now, we know that :
1. Maximum projectile occurs in Ø=45º, & T will be :
T=(1/g)Vosin(45º)+(1/g)√[Vo^2sin(45º)^2]. Now, let's calculate
the integral : S=∫√[Vo^2*cos(45)^2+(Vo*sin(45º)-gt)^2]dt
S=∫√[Vo^2+(gt)^2-2Vo*gt]dt. This sort of integral can be find in
any tables of integrals.

Alon.  

Calculus

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Alon Mandes

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Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .

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