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# how can I solve this problem..

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youn chan choi on 20 Apr 2021
Edited: youn chan choi on 21 Apr 2021
syms theta0
th = -y+cos(theta0)*x^2-g*x^2/(2*v0^2*sin(theta0)^2) == 0;
theta0 = [0 pi];
stheta0 = fzero(th,theta0);
I want to get two roots for theta0 but when I erased 'theta0 = [0 pi]' , it was too complicated.
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### Answers (1)

John D'Errico on 20 Apr 2021
Edited: John D'Errico on 20 Apr 2021
Um, what do you expect?
syms theta0
x=90; g=9.81; v0=30; y0=1.8; y=1;
th = -y+tan(theta0)*x-g*x^2/(2*v0^2*cos(theta0)^2)+y0;
pretty(th)
8829 4 90 tan(theta0) - ---------------- + - 2 5 200 cos(theta0)
Do you expect an analytical solution, something nice and simple and very pretty?
thsol = solve(th)
thsol = You should be able to effectively reduce this to effectively a 4th degree polynomial in terms of sin(theta0), so 4 roots.
vpa(thsol)
ans = There are probably infinitely many solutions since trig functions are periodic. I don't see how much better than that you can rationally expect? Honestly, I'd say you are pretty lucky, in that solutions are easily generated\. Just pick the one that makes sense to you.
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