A cannon (at the origin) shoots a ball at an angle [tex] \theta [/tex] above the horizontal ground. Neglect air resistance. Let [tex] r(t) [/tex] denote the ball's distance from the cannon. What is the largest possible value of [tex] \theta [/tex] if [tex] r(t) [/tex] is to increase throughout the ball's flight? Taylor, Classical Mechanics problem 1.40(adsbygoogle = window.adsbygoogle || []).push({});

Alright, so I have the equations for the components of the distance vector:

[tex] x = \cos(\theta) \cdot v_{o} \cdot t [/tex]

[tex] y = \cos(\theta) \cdot v_{o} \cdot t - \frac{g^2}{2} \cdot t^2 [/tex]

Next I create the distance vector, take its magnitude([tex] \sqrt{x^2 + y^2} [/tex]), and square it to get:

[tex] v_{o}^2 \cdot t^2 + \frac{g^2}{4} \cdot t^4 - g \cdot t^3 \cdot v_{o} \cdot \sin(\theta) [/tex]

My next step would be to take the derivative of the function (with respect to time) and set the condition that time is greater than zero. This would allow me to keep the derivative monotonic. With this in mind I would find where the derivative is equal to zero and solve for time by using the quadratic formula. Then I would take the discriminant, set it to be greater than zero, and find solve for theta.

[tex] \frac{d(r^2)}{dt} = t(2 \cdot v_{o}^2 + g^2 \cdot t^2 - 3g \cdot t \cdot v_{o} \cdot \sin(\theta)) [/tex]

However, when I do this process the discriminant gives me the condition that [tex] \theta [/tex] is to be greater than 70.5 degrees. This is obviously wrong but no matter how many times I've looked at it today I can't find my mistake. Anybody out there willing to help me figure out what I'm doing wrong?

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# Homework Help: Projectile Motion of a cannon

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