Calculating time from acceleration and distance travelled

Question

I have an interesting question that has me stumped. If an object is free falling (so accelerating at 9.8m/s^2), by gravity, and it traveled a distance of 26cm, how long did that take? The initial velocity is 0 m/s and I don’t know the final velocity. I’m doing a reaction time test but can’t work out the actually time. Thanks in advance to all who participated. MathsCuriosity

Answer 1

During a reaction time test, an object is released into free fall motion. So the Equation of Motion of the object is $\ddot{x}=g$, where $x$ is the distance traveled by the object according to a ruler held next to the experiment. Since the force is constant, we simply integrate the EoM to get the result $x=\frac{g}{2}t^2+v_0t+x_0=\frac{g}{2}t^2$, where $v_0$ and $x_0$ are the initial velocity and initial position, respectively, and are both $0$ here.

So since the distance traveled is $x=26\text{cm}$ and $g=9.8\text{ms}^{-2}$, and so the time elapsed from the time that the object was released, to the end of the 26cm that the object traveled when it was caught is $t=\sqrt{\frac{2x}{g}}=\sqrt{\frac{2*26cm}{9.8}}=.230272\text{s}\approx 230\text{ milliseconds}$.

When solving for $t$, in the equation $x=\frac{g}{2}t^2$, we eventually have to take the square root. Often we are asked to find the positive and negative root (for example the quadratic equation is generally written with a $\pm$ sign). Here the negative solution would correspond to what would correspond to before the object was even dropped, and so is nonsense.