It seemed reasonable to me when I started solving it. It is uniform, and I'd assumed that all the force the person doing the chin-up exerts is solely confined to the first half of the chin-up.
Although now that you've pointed it out, the athlete in question doesn't really let go of the bar, so...
Oh, thank you so much! I didn't quite make the connection between the "uniform acceleration" and the "taking the same time" parts. It makes sense now. I can replicate the successful solution now without a hitch, (in fact, if ##(\frac{F}{m} - g)\times\frac{t_{1}^{2}}{2} = 0.15##, I get the...
This is a fairly straightforward problem. I'll just post the way that I'd solved it: for some reason I'm getting the wrong answer.
$$m\times a_{0} = F - w$$
$$a_{0} = \frac{F}{m} - g$$
$$v_{0} = (\frac{F}{m} - g) \times t$$
$$y_{0} = (\frac{F}{m} - g) \times \frac{t^{2}}{2}$$
This gives us the...