(N.B: SUVAT questions are made up of 5 parts: displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t).)
I've got a question that is seemingly impossible, though it shouldn't be since it was given out as homework:
My plan of action was to find the point where the ball's velocity was 0 (93.06m) then treat the question as a "drop the ball" question, though I ran into an issue.
(useful information: $s = 93.06$, $u = 0$, $a = -9.8$)
When calculating the speed at which it hit the floor, the equation I ended up with (from $v=±\sqrt{2as+u^2}$) was $v=±\sqrt{-1834.56}$, which is imaginary and therefore not an answer to a SUVAT question.
I can't see any way to avoid this, as the acceleration will always be a negative number.
What am I doing wrong?
Remember, $s$ denotes displacement at the end relative to the beginning, not the other way round. Since the acceleration is negative, $s$ should be negative, i.e. $s=-93.06$. Another way to look at it is that $s=s_0+ut+\frac12at^2$ and $v^2=u^2+2a(s-s_0)$ with $s_0=93.06$, and the ball hits the ground at $s=0$ whence $s-s_0=-s_0=-93.06$. Another way to solve the problem is energy conservation, which again explains why the squared speed increases by $2\times 80\times 9.8$ from its itnial value of $16^2$.