Answer:
![s=vt-(1)/(2)gt^2](https://img.qammunity.org/2020/formulas/physics/middle-school/rvv8voybsmnrc1oi5o96nhwtcp4p40jdmp.png)
Step-by-step explanation:
We could use the following suvat equation:
![s=vt-(1)/(2)gt^2](https://img.qammunity.org/2020/formulas/physics/middle-school/rvv8voybsmnrc1oi5o96nhwtcp4p40jdmp.png)
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
![g=-9.8 m/s^2](https://img.qammunity.org/2020/formulas/physics/high-school/rbj9888tfpdney5og7fmxd9wulotvtc2g1.png)
And the coin reaches the water when
t = 1.3 s
Substituting these data, we can find v:
![v=(s)/(t)+(1)/(2)gt=-(1.2)/(1.3)+(1)/(2)(-9.8)(1.3)=-7.3 m/s](https://img.qammunity.org/2020/formulas/physics/middle-school/bbv5ca4oshx3aecmt0jswphw7wjd46j1gu.png)
where the negative sign means the direction is downward.