"You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 9.00 s after it was thrown. What is the speed of the rock just before it reaches the water 21.0 m below the point where the rock left your hand

  48.54 m/s

Explanation:

If the rock takes 9 seconds to reach your position after being thrown, it reaches its maximum height in 4.5 seconds.

The height the rock reaches above your position is ...

  h = (1/2)gt^2 = (4.9 m/s^2)(4.5 s)^2 = 99.225 m

This height is an additional 21 m above the water, so the maximum height above the water is ...

  99.225 m +21.0 m = 120.225 m

The velocity (v) achieved when falling from this distance is found from ...

  v^2 = 2gh

  v = √(2(9.8)(120.225)) = √2356.41 ≈ 48.543 . . . . m/s

The speed of the rock when it hits the water is about 48.54 m/s.

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