The problem is as follows:
At a laboratory a sphere is launched from the ground vertically upwards. The technician in charge measures the time it takes for the ball to reach the launching site again and labels this time as $\tau$ in his computer. Find the elapsed time in terms of $\tau$ which is between the launching start until the sphere is going downwards at a height which is $50\%$ of its maximum height. Use $g=10\,\frac{m}{s^2}$
The alternatives given on my book are:
$\begin{array}{ll} 1.&0.50\tau\\ 2.&0.60\tau\\ 3.&0.65\tau\\ 4.&0.75\tau\\ 5.&0.85\tau\\ \end{array}$
I attempted to solve this problem but the only thing which I could come up with was to establish the known motion equations for free fall.
$y=y_{o}+v_{oy}t-\frac{1}{2}gt^2$
Since in this case is thrown vertically but from ground I am assuming that x_{o}=0, which would make the above equation into:
$y=v_{oy}t-\frac{1}{2}gt^2$
Now the equation states that for the allotted time between launching start and reaching that site again would become into:
$y(t)=v_{oy}t-\frac{1}{2}gt^2$
$0=v_{oy}\tau-\frac{1}{2}g\tau^2$
From which the initial speed can be calculated:
$v_{oy}\tau=\frac{1}{2}g\tau^2$
$v_{oy}=\frac{\frac{1}{2}g\tau^2}{\tau}$
Since I'm treating $\tau$ as a constant there shouldn't be a problem by cancelling this factor:
$v_{oy}=\frac{1}{2}g\tau$
Therefore the initial equation becomes into:
$y(t)=\left(\frac{1}{2}g\tau\right) t-\frac{1}{2}gt^2$
Now I'm assuming that the time that it takes to get to the top would be half of the flying time so $t=\frac{\tau}{2}$ and that would be the height.
$h=y\left(\frac{\tau}{2}\right)=\left(\frac{1}{2}g\tau\right)\frac{\tau}{2}-\frac{1}{2}g\left(\frac{\tau}{2} \right )^2$
$h=\frac{\tau^{2} g}{4}-\frac{\tau^{2}g}{8}=\frac{\tau^{2} g}{8}$
Now to get $50\%$ of that height would be:
$\frac{h}{2}=\frac{\tau^{2} g}{16}$
So all that there is just to sum the time that will take for the object to travel that distance to the ground. I understood this part as finding the time from the beginning until the sphere reaches that height and adding that time to $\frac{\tau}{2}$.
Returning to the original equation the time can be obtained from:
$y(t)=\left(\frac{1}{2}g\tau\right) t-\frac{1}{2}gt^2$
$y(t)=\left(\frac{1}{2}g\tau\right) t-\frac{1}{2}gt^2=\frac{\tau^{2} g}{16}$
So solving for $t$ I can obtain the elapsed time until the sphere gets to half the height.
Rearranging the equation becomes into:
$\frac{1}{2}gt^2-\left(\frac{1}{2}g\tau\right) t+\frac{\tau^{2} g}{16}=0$
$8gt^2-8g\tau t+\tau^{2} g=0$
$t=\frac{8\tau\pm \sqrt{64\tau^2-32\tau^{2}}}{16}=\frac{8\tau\pm \sqrt{32\tau^{2}}}{16}$
$t=\frac{8\tau\pm 4\sqrt{2}\tau}{16}$
$t=\frac{2\tau\pm \sqrt{2}\tau}{4}$
For this part the logic tells to use positive time, but both alternatives in the quadratic equation render positive results:
But I'm assuming that the second alternative gives the result directly in other words when the sphere goes up and down to the $50\%$ of the height. Am I right in this assumption?
$t\approx \frac{2+1.4142}{4}\tau$
Therefore:
$t\approx \frac{2+1.4142}{4}\tau\approx 0.853553 \tau$
and the answer would be $0.853553 \tau$ or the fifth alternative.
Now going on the other route and using the first result would mean to add $\frac{\tau}{2}$ to that result. I mean the other alternate result given by the quadratic equation.
This is described as follows:
$t=\frac{\tau}{2}+\frac{2\tau - \sqrt{2}\tau}{4}$
$t=\frac{2\tau+2\tau-\sqrt{2}\tau}{4}=\frac{4\tau-\sqrt{2}\tau}{4}\approx 0.6464 \tau$
Which in turn would be $0.65 \tau$ as it appears in one of the alternatives.
But apparently my book states that the right answer is the first one. In other words the one which I found earlier. But I'm still stuck at why my second assumption did not yielded the correct result?.
Can somebody please follow my steps and tell me if did I missunderstood something and what mistake did I made?. I need guidance not only in the mathematical part but in the physics concepts as well as I'm not sure if the way how I'm using the algebraic part is correct or not. Overall can somebody help me with this?
A more direct approach could be useful here. The time the sphere takes to fall from its top position to the ground is $\tau/2$, hence: $$ h={1\over2}g\left({\tau\over2}\right)^2. $$ If $t'$ is the time the sphere takes to fall from top to $h/2$ we have: $$ {h\over2}={1\over2}gt'^2. $$ Comparison gives: $$ t'={\tau\over2\sqrt2} $$ and the time taken from start to arrive at the same position is then: $$ t={\tau\over2}+t'={2+\sqrt2\over4}\tau. $$ Your computation is correct: both times you found correspond to the time taken to arrive at $h/2$ from start, the first time when rising and the second time when falling. The error is in the assumption that the difference between these two times is $\tau/2$: the true difference is of course twice the time $t'$ computed above, that is $\tau/\sqrt2$.