Extraneous solutions from equation of motions

Question

Suppose upwards vector is positive and downwards vector is negative.

A ball is released from a tower of height $100 \, \rm m$ ($g=10 \, \rm m/s^2$). Find its final velocity

Using the equation of motion $v^2=u^2+2as$, $v^2=\left(0^2+2\times(-10) \times(-100)\right) \rm m^2/s^2$ or $v^2=2000 \, \rm m^2/s^2$, which means $v=\pm\sqrt{2000} \, {\rm m/s}=\pm20\sqrt{5} \, {\rm m/s}$.

Now it is pretty obvious that $v=-20\sqrt{5} \, \rm m/s$, but why do we get the extraneous solution? What does it actually mean?

Answer 1

If you look at how the equation is derived, it will help.

For simplicity, let's forget about the initial velocity, $u$, as it is zero.

The velocity after time $t$ is:

$$v = at \quad (-10t \,\,\,\, \text{in this case}).$$

There is a linear increase in velocity from $0$ to $at$ (constant acceleration), so the average velocity during this time is $1/2 \, at$, and, therefore, the distance travelled in time $t$ is:

$$s = \frac{1}{2} at \cdot t.$$

This gives us a $t^2$ term, which would give us a $v^2$ term if we substitute for $t$ in our original equation $v=at$. (After substitution, it gives us $v^2=2as$.)

So, think about the $t^2$ term. We get the same value if $t$ is positive or negative. So, what does "negative $t$" mean? Well, minus $10 \ \text{s}$ means $10 \ \text{s}$ before the ball is released.

But the simple equation of motion does not account for the holding and releasing of the ball - it describes a continuous motion under constant acceleration, not necessarily starting at the point where the ball is at zero velocity at the top of the tower.
In this continuous motion, negative time is when the ball would be moving upwards (for example, having been thrown upwards) towards the top of the tower, decelerating so that it has zero velocity when it reaches the top of the tower. So this upwards (positive) velocity is the other solution to the equation.