- #1

brotherbobby

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- Homework Statement
- A cannon sits atop on a plane at a height ##h## above the ground and fires a shell with an initial velocity ##v_0##. At what angle ##\theta_0## must it fire the shell so as to attain maximum range ##R_{\text{max}}## along the ground?

Ans : ##\boxed{\boldsymbol{\theta_0 = \sin^{-1} \frac{v_0}{\sqrt{2(v_0^2+gh)}}}}##

- Relevant Equations
- (1) Equation of motion of the shell in the vertical direction : ##y = h+ v_0 \sin\theta_0 t-\frac{1}{2} gt^2##.

(2) Equation of motion in the horizontal direction : ##x = v \cos \theta_0 t## for any time ##t##.

My approach is going to be along the following lines : (1) Use the equation of motion in the ##x## direction to express the total time of flight ##T## as a function of the range and initial angle ##T = T(R, \theta_0)##. (2) Plug this value of time into the equation of motion in the ##y## direction to express the height ##h## as a function of range, gravity and initial angle ##h = h(R,g, \theta_0)##. (3) Express the range ##R## as a function of height, initial angle, gravity and initial velocity : ##R = R(h, \theta_0, v_0, g)##. (4) Differentiate the equation of ##R## with respect to ##\theta_0## and set it equal to zero and solve the resulting equation to express the initial angle as a function of height and initial velocity : ##\boxed {\theta_0 = \theta_0(v_0, h, g)}##. This should be the answer to the problem.

**My attempt :**

Range ##R = v_0 \cos\theta_0 T## where ##T## is the time period of flight. Hence ##T = \frac{R}{v_0 \cos\theta_0}##. Again, the height ##-h = v_0\sin\theta_0 T - \frac{1}{2}gT^2 \Rightarrow -h = R\tan\theta_0 - \left( \frac{gR^2}{2v_0^2}\right) \frac{1}{\cos\theta_0^2}##.

Simplifying and rearranging, this leads to a quadratic equation in ##R## : $$R^2 - \frac{v_0^2 \sin 2\theta_0}{g}R-\frac{v_0^2 h}{g}(1+\cos 2\theta_0)=0$$ Using the quadratic formula, we solve for ##R = \frac{v_0^2 \sin 2\theta_0}{2g}+\frac{1}{2g}\sqrt{\frac{v_0^4}{2}(1-\cos 4\theta_0)+4v_0^2 gh(1+\cos 2\theta_0)}##. Differentiating this with respect to ##\theta_0##, we obtain $$\frac{dR}{d\theta_0} = \frac{v_0^2 \cos 2\theta_0}{g}+ \frac{1}{2g\sqrt{\mathscr{E}}}(2v_0^4 \sin 4\theta_0-8v_0^2 gh \sin 2\theta_0)$$ where ##\mathscr E## stands for the expression inside the square root sign in the equation above. Setting this derivative to 0 for maximum range ##R_{\text{max}}## and writing the expression for ##\mathscr {E}##, we get the following equation: $$\cos 2\theta_0+\frac{v_0^2 \sin 4\theta_0-4gh \sin 2\theta_0}{\sqrt{\frac{v_0^2}{4}(1-\cos 4\theta_0)+4v_0^2gh(1+\cos 2\theta_0)}}=0\;\; \left(\frac{v_0^2}{g}\neq 0 \right)$$

The problem is, upon squaring and simplifying the above equation in order to get rid of the square root, this leads to a near impossible equation to solve in trigonometry after several lines of slightly tedious algebra: $$\frac{3}{8} \cos 8\theta_0-3v_0^2 gh \cos 6\theta_0 - (8g^2h^2-2v_0^2gh)\cos 4\theta_0 - 3v_0^2gh\cos 2\theta_0 - \left(\frac{3}{8} v_0^4 - 2v_0^2gh-8g^2h^2 \right) = 0$$

I am supposed to solve this equation in order to express ##\theta_0 = \theta_0(h, g, v_0)##, but I wonder if the above equation be solved.

**A hint or help would be welcome.**

For instance, do we have softwares (online or otherwise) that can solve trigonometric equations?

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