Answer
$h \approx 73.39 m$
Work Step by Step
For the whole distance
$h = \frac{gt^{2}}{2}$
The rock travels 100% - 45% = 55% of the whole distance before the last second, therefore
$0.55h = \frac{g(t-1)^2}{2}$
Plug in $h$ in the second equation
$0.55\times\frac{gt^{2}}{2}=\frac{g(t-1)^2}{2}$
$0.55t^2=(t-1)^2$
$t_{1} \approx 0.574s$
$t_{2} \approx 3.87s$
Only $t_{2}$ is valid since the rock traveled for at least 1 sec. Finally, total height
$h\approx\frac{g\times3.87^{2}}{2}\approx73.39m$
