Question

notice a water ballon fall past your classroom window. You estimate that it took the balloon about t seconds to fall the length of the window and that the window is about y meters high. Suppose the balloon started from rest. Approximately how high above the top of the window was it released?

Answer 1

Let's denote the distance between the top of the window and the start point of balloon by $h$. We shall use the law for uniformly accelerated motion. Let's assume that the balloon travelled this distance $h$ in some time $\tau$ (we don't know it). Then it travelled distance $h + y$ in time ${\tau + t}$ (we know it from the conditions of the problem - it travelled the distance $y$ in time $t$ ). The law for uniformly accelerated motion give us two equation for these cases (we use that the balloon started from rest thus ${v_0} = 0$ and let the initial point of the coordinate system at the initial position of the ballon)

$$\left\{ \begin{array}{l}h = \frac{{g{\tau ^2}}}{2}\\h + y = \frac{{g{{(\tau + t)}^2}}}{2}\end{array} \right.$$

Using the first equation we get

$$\tau = \sqrt {\frac{{2h}}{g}}$$

Let's substitute this to the second equation and simplify it

$$h + y = \frac{g}{2}{(\sqrt {\frac{{2h}}{g}} + t)^2} = \frac{g}{2}(\frac{{2h}}{g} + 2t\sqrt {\frac{{2h}}{g}} + {t^2}) = h + \frac{{g{t^2}}}{2} + t\sqrt {2hg}$$

Transfer all terms except the last one from the right to the left side

$$t\sqrt {2hg} = y - \frac{{g{t^2}}}{2}$$

squaring

$$2hg{t^2} = {(y - \frac{{g{t^2}}}{2})^2}$$

and expressing $h$

$$h = \frac{{{{(y - \frac{{g{t^2}}}{2})}^2}}}{{2g{t^2}}}$$

This is the answer

P.S. It can be expressed in different (but equivalent) forms

$$h = \frac{{{{(y - \frac{{g{t^2}}}{2})}^2}}}{{2g{t^2}}} = \frac{{{{(2y - g{t^2})}^2}}}{{8g{t^2}}} = \frac{{g{t^2}}}{8} - \frac{y}{2} + \frac{{{y^2}}}{{2g{t^2}}}$$