# Ask Uncle Colin: A load of parabolics

Dear Uncle Colin,

One of my students recently attempted the following question: “At time $t=0$ particle is projected upwards with a speed of 10.5m/s from a point 10m above the ground. It hits the ground with a speed of 17.5m/s at time $T$. Find $T$.” They used the equation $s = vt - \frac 12 at^2$ and found $T = \frac{5}{7}$ and $T=\frac{20}{7}$ - but the other answers only give the latter of those. Where has the phantom solution come from?

- Suddenly Upset, Vexing Additional Time

Hello, SUVAT, and thank you for your message!

There’s a graphic doing the rounds at the moment showing the difference between how maths is actually done (following wrong paths, backtracking, correcting errors) against how it’s usually presented (we do this then this then this and boom! answer). This question is a prime example of that: two teachers scratched their heads for several hours going back and forth, eventually stumbled on the answer, and even then weren’t sure it was right until having slept on it. All the same, I’m going to present just the reasoning we ended up with rather than all the wrong paths.

Most students would, I think, use one of the other SUVAT ((Hey! That’s your name!)) equations to solve this one - $s = \frac{t(u+v)}{2}$ is probably favourite - the only one that absolutely doesn’t work is $v^2 = u^2 + 2as$, because it doesn’t have time in it at all.

There’s nothing obviously ‘wrong’ with using the equation $s = vt - \frac{1}{2}at^2$ with the values given in the question: $-10 = -17.5T + 4.9T^2$, which rearranges into a nicer quadratic as $49T^2 - 175T + 100 = (7T-20)(7T-5) = 0$. Two solutions - but only one is valid. And, looking at the ball’s trajectory, $T=\frac 57$ isn’t a particularly interesting point on the journey - the ball is 15m above the ground at that point. So why is it coming up as an answer?

Short answer: because we’re measuring from the wrong end.

Longer answer: the equation $s = vt - \frac{1}{2}at^2$ tells you the displacement you underwent in the last $t$ seconds given you travelled with a constant acceleration $a$ and ended up at velocity $v$. If you state the displacement and solve for $t$, you find every possible time you could have travelled for, given that you were eventually displaced by $s$. In this case, $\frac 57$ seconds *before landing*, the particle was 10m above the ground, but heading downwards - and that’s where the spurious solution came from.

This is similar, in some respects, to solving $v^2 = u^2 + 2as$ for $v$ or $u$ and deciding between two possible velocities - only it’s less obvious that you have different answers, because $v$ and $u$ can be thought of as speeds if you’re waving your hands around enough.

Hope that clears it up a bit, SUVAT!

- Uncle Colin

* Thanks to @srcav for discussing this problem with me!