A stone is dropped into water from a bridge 44.1 m above the water. Another stone is thrown vertically downward 1 s later. Both strike the water simultaneously. What was the initial speed of the second stone?

Explanation

To solve the problem step by step, we will follow the kinematic equations of motion.

### Step 1: Calculate the time taken by the first stone to hit the water.
The first stone is dropped from a height of 44.1 m. We can use the second equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
Where:
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