So I've been given the following equation of motion of a particle:
$$\ddot{x} = g - {2gx\over a} \sin{\theta}$$
And have been told to:
Find an expression in terms of $\space \theta \space$, for $\space {1\over 2}v^2 \space$, where $\space v \space$ is the speed of the particle.
My instincts told me to look at the following equality:
$$\ddot{x} = v{dv\over dx}= g - {2gx\over a} \sin{\theta}$$
Integrating both sides w.r.t. $\space x \space$ gives:
$$\int v \space dv = {1\over 2} v^2 = gx -{gx^2 \over a}\sin{\theta} + C$$
At $\space v=0, \space x=0 \space$ meaning $\space C=0$. So:
$${gx^2 \over a}\sin{\theta} = gx - {1\over 2} v^2$$
Then:
$$\sin{\theta} = {a\over x} - {a\over 2gx^2}v^2$$
Then finally:
$$\theta = \arcsin \huge( \normalsize{a\over x} - {a\over 2gx^2}v^2 \huge)$$
Is this what the question meant by "Find an expression in terms of $\space \theta$"?
Or is it asking something completely different?