# Find the height of a building given that it takes an object in freefall $1$ second to reach the ground from half the building's height

An object is dropped from a building of height $h\,\mathrm m$. The object falls half the total distance to the ground during the last $1.00\,\mathrm s$ of its fall. Find the height of the building. Take the downward direction to be negative and $9.80\,\frac{\mathrm m}{\mathrm s^2}$ to be the magnitude of the acceleration due to gravity.

Let $t,y,y_0,v_y,{v_y}_0$ respectively denote the time (in seconds), vertical displacement (in meters), the initial displacement, the vertical velocity (in meters per second), the initial velocity of the object. Let $a$ denote the acceleration's magnitude as above.

I've split up the drop into two time intervals, $0\le t\le T$ and $T\le t\le T+1$, where $T$ is the time it takes for the object to fall half the building's height from when it is dropped.

In the first time interval, I see that

$$\begin{cases}y_0=0\\{v_y}_0=0\end{cases}\implies\begin{cases}v_y=-at\\y=-\frac a2t^2\end{cases}$$

At the end of the first time interval, half the building's height is traversed, so that

$$-\dfrac h2=-\frac a2T^2\implies h=aT^2\quad(*)$$

In the second time interval, it seems clear to me that

$$\begin{cases}y_0=-\frac h2\\{v_y}_0=-aT\end{cases}\implies y=-\frac h2-aTt-\frac a2t^2$$

so that when the object hits the ground at $t=T+1$, I have

$$-h=-\frac h2-aT(T+1)-\frac a2(T+1)^2\implies h=a(3T^2+4T+1)\quad(**)$$

Equations $(*)$ and $(**)$ suggest that

$$3T^2+4T+1=T^2\implies T\approx-1.71,T\approx-0.293$$

which doesn't make sense to me. The time $T$ that elapses throughout the object's fall should be positive.

The solution is supposed to be

$h\approx57.1$ meters

which according to $(*)$ should correspond to $T\approx2.41$ seconds; on the other hand, $(**)$ would suggest $T\approx0.766$ seconds.

I'm fairly certain $(**)$ is incorrect, but I don't know what mistake I made in its derivation.

In your equation$$-h=-\frac h2-aT(T+1)-\frac a2(T+1)^2\implies h=a(3T^2+4T+1)\quad(**)$$You plugged in $(T+1)$ for $t$, which is not correct. You need to plug in the duration of the second half of the fall, which was $t=1$. So, the formula should be$$-h=-\frac h2-aT(1)-\frac a21^2\implies h=a(2T+1)\quad(**)$$Solve this to get the actual answer.
• You left a factor of $T+1$ in your answer. I've gone ahead and edited. Thanks again! – user170231 Aug 18 '18 at 2:43