I just took a quick look at this test today. Some interesting question, and parts of Q14 are definitely pretty difficult.
Q14d has an interesting kink that I don't think any of the current solutions have captured. The question doesn't state that
, though most of the solutions seem to assume that. I think we should take
, as there is nothing in the question to suggest that downward projection of the particle is not allowed (and the other two quadrants are uninteresting as they are identical except for leftward motion instead of rightward).
Why this is interesting is that the solution to
 < 8/9)
is
< \theta < \sin^{-1}(\sqrt{8/9}))
However this is not correct, as for negative
(downward projected particle) the distance from the origin is an increasing function of t for all
.
The trick here is that the resultant equation,
 t + 2 V^2)
must be positive definite (>0) for the case of positive
, as the axis of symmetry (-b/(2a)) is positive and
.
However, for the case of
the axis of symmetry is negative, so it's not necessary for the quadratic to be positive definite (>0 for all t), as it only needs to be positive for
. And all that's required to prove that is to note that for
that all the coefficients are positive and therefore the quadratic is positive for all
This gives the overall solution
Hope that helps.
because i'm preparing for the worst since my internals look kinda dogshit rn
