The velocity of the stone when it hits the water, i.e., the final velocity of the stone, v=68.6 m/s
The acceleration of the stone, a=9.8 m/s²
As the stone was dropped (not thrown), the initial velocity of the stone, u=0 m/s
From one of the equations of the motion,
[tex]v=u+at[/tex]
Where t is the time interval in which the stone hits the water.
On rearranging the above equation,
[tex]t=\frac{v-u}{a}[/tex]
On substituting the known values in the above equation,
[tex]\begin{gathered} t=\frac{68.6-0}{98} \\ =0.7\text{ s} \end{gathered}[/tex]
Therefore the stone hits the water 0.7 seconds after it is dropped.
