(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?

  • $\begingroup$ It's unclear how you got -1834.56; from your numbers (correcting the sign of $s$) I get slightly less than 1824. I get exactly 1824 using the same formula from the starting point, which I think is easier in this case. $\endgroup$ – David K Mar 16 '20 at 13:07
  • $\begingroup$ You really don’t need to find the ball’s maximum height. By symmetry and conservation of energy, it will have the same speed as its initial speed once it returns to the same altitude, but directed downwards. $\endgroup$ – amd Mar 16 '20 at 18:42
  • $\begingroup$ @DavidK Yeah, I didn't have a negative for my displacement, hence the issue. $\endgroup$ – Corsaka Mar 17 '20 at 9:06

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 $16^2$.


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