Ytukyg

Can the equation of the form
$$sin(omega t) + at^2+bt=N$$
where $a,b,N in mathbb{R}$ be solved analytically for $a ne 0$?

It, for example, arises when the rotating object falls down subject to gravity. Solving such equation can allow to compute the time of the object touchdown.

Your equation is a transcendental equation. It can not be exactly solved but you can find the roots using numerical methods like Newton's method. However if you are still interested in having a closed form solution, you may use Bhaskara's approximation for domain $[-frac {π}{2},frac {π}{2}]$. For example, consider another related equation,
$$cos omega t + t^2 + bt - N = 0$$
From Bhaskara's approximation formula,
$$cos omega t ≈ frac {π^2 - 4omega^2 t^2}{π^2 + omega^2 t^2}$$
The equation is reduced to a general quartic,
$$t^4+bt^3+(T^2 - (4+N))t^2+T^2bt+T^2(1-N)=0$$
Where $T=frac {π}{omega}$This can be solved using Ferrari's method (https://proofwiki.org/wiki/Ferrari%27s_Method). Although the computational cost is high and the solution may turn really ugly, yet it will give you much required insight into the dependence of the time of flight of the object on various parameters.

In the same spirit as Awe Kumar Jha, considering $$sin (omega t) + at^2+bt=N$$ let
$$x=omega tqquad alpha=frac a{omega^2}qquadbeta=frac b{omega}$$ to make the equation
$$sin(x)+alpha x^2+beta x=N$$ and use Bhaskara I's sine approximation formula
$$sin(x) simeq frac{16 (pi -x) x}{5 pi ^2-4 (pi -x) x}qquad (0leq xleqpi)$$
$$4 alpha x^4+ 4( beta - pi alpha )x^3+ left(5 pi ^2 alpha -4 pi beta -4 N-16right)x^2+pi (5 pi beta +4
N+16)x-5 pi ^2 N=0tag 1$$ which, as said,can be solved using Ferrari's method (this would be quite tedious).

You can use another approximation (from far, not as good asBhaskara I's one ) minimizing with respect to $k$
$$int_0^pi left[sin(x)-k (pi-x)xright]^2,dx=frac{pi ^5 }{30}k^2-8 k+frac{pi }{2}$$ to get $color{blue}{k=frac{120}{pi ^5}}$ and then $sin(x)simeq k(pi-x)x$. This leads to
$$ (alpha -k)x^2+ (beta +pi k)x-N=0tag 2$$ which is simpler for sure. Using its solution as $x_0$ you can start Newton iterations to polish the root accorging to
$$x_{n+1}=x_n-frac{sin (x_n)+alpha x_n^2+beta x_n-N}{ cos (x_n)+2 alpha x_n+beta}$$

Let us try with $alpha=0.4$, $beta=0.6$, $N=2.345$. Solving $(1)$ gives $x^{(1)}_0 =1.26126$ while solving $(2)$ gives $x^{(2)}_0=1.27312$. Now, the iterates of Newton method starting with $x_0=x^{(2)}_0$
$$left(
begin{array}{cc}
n & x_n \
0 & 1.273117520 \
1 & 1.260970996 \
2 & 1.260976964
end{array}
right)$$ which is the solution for ten significant figures.

Edit

To stay with a quadratic equation, we could be even more empirical (not respecting the values at end points) and write
$$sin(x) simeq a_0+a_1x+a_2x^2$$ The same minimization method would lead to
$$a_0=frac{12 left(pi ^2-10right)}{pi ^3}qquad a_1=-frac{60 left(pi ^2-12right)}{pi ^4}qquad a_2=frac{60 left(pi ^2-12right)}{pi ^5}$$ which would lead to $x^{(3)}_0=1.26757$.