Question

A stone is dropped from the top of a cliff and is found to travel $44.1m$diving the last second before it reacts the ground. What is the height of the cliff? $g=9.8m/s^2$.

Answer 1

$44.1=H$-(distance travled in ($\sqrt{\frac{2H}{g}}-1s$)

Time of flight $=\sqrt{\frac{2H}{g}}$

Distance traveled in $\left(\sqrt{\frac{2H}{g}}\right)-1$=$\frac{1}{2}g$

($\sqrt{\frac{2H}{g}}-1{)}^2$

$=\frac{g}{2}(\frac{2H}{g}+1-2\sqrt{\frac{2H}{g}})$

$=H+\frac{g}{2}-\sqrt{2gH}$

$44.1=\sqrt{2gH}-\frac{g}{2}$

$=\sqrt{2gH}-4.9$

$\sqrt{2gH}=49$

$2\times 9.8\times H=49\times 49$

$H=\frac{49\times 49\times 10}{9.8\times 2}$

$H=\frac{490}{4}=122.5m$