Maximum Trajectory Length

Classical Mechanics Level 4

A ball is thrown at speed $v$ from zero height on level ground. Let $\theta$ be the angle at which the ball should be thrown so that the length of the trajectory is maximum. Find $\theta$ (in degrees).

(Use a calculator or any other computing device in the last step to calculate $\theta$ )

Ignore all viscous and drag forces.

The answer is 56.5.

3 solutions

Utkarsh Tiwari
Mar 14, 2016

can have the solution here.... solution ..P.S. i didnt copy it from here...

Oh so it wanted the arc length not the maximum horizontal distance

Frederick Cooke - 5 years, 3 months ago

Mind blowing soln

Dhruv Aggarwal - 5 years, 3 months ago

The equation I got for the total length of the parabola is: (x is the angle) $f(x,v)=\sqrt{\left(\frac{2v^2\sin(x)\cos(x)}{9.81}\right)^{2}+4\left(\frac{v^2\sin^2(x)}{19.62^2}\right)^{2}}+\frac{\left(\frac{2v^2\sin(x)\cos(x)}{9.81}\right)^{2}}{2v^2\sin^2(x)}\cdot\sinh^{-1}\left(\frac{2v^2\sin^2(x)}{\frac{2v^2\sin(x)\cos(x)}{9.81}}\right)$

I probably messed up somewhere...

A Former Brilliant Member - 5 years, 2 months ago

Ak Gh
May 13, 2016

Must integrate, S=arc length= $\int ds=\int \sqrt{dx^2+dy^2} =\int \sqrt{(dx/dt)^2+(dy/dt)^2} dt$ .

i.e., You must integrate $\int_0^S ds=\int_0^T \sqrt{(v\cos(\theta))^2+(v\sin(\theta)-gt)^2} dt$ ., T is the total time of flight

So you get S( $\theta)$ . Now maximize S( $\theta)$ by using calculus. That is how I did.

If there is any other simpler solution, I'll be pleased to know. (perhaps using calculus of variations?)

Shashank Goel
Mar 19, 2016

I got the length of the trajectory as (u^2)/(2g) (2sinx+cos^2(x) ln((1+sinx)/(1-sinx))) and maximized it using wolfram Got x=0.985515 radians

Maximum Trajectory Length

The answer is 56.5.

3 solutions

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