Answer:
[tex]s=vt-\frac{1}{2}gt^2[/tex]
Explanation:
We could use the following suvat equation:
[tex]s=vt-\frac{1}{2}gt^2[/tex]
where
s is the vertical displacement of the coin
v is its final velocity, when it hits the water
t is the time
g is the acceleration of gravity
Taking upward as positive direction, in this problem we have:
s = -1.2 m
[tex]g=-9.8 m/s^2[/tex]
And the coin reaches the water when
t = 1.3 s
Substituting these data, we can find v:
[tex]v=\frac{s}{t}+\frac{1}{2}gt=-\frac{1.2}{1.3}+\frac{1}{2}(-9.8)(1.3)=-7.3 m/s[/tex]
where the negative sign means the direction is downward.
