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 23.0 m below the point where the rock left your hand

Explanation:

After 9 seconds displacement is zero

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = ?

        Acceleration, a = -9.81 m/s²  

        Time, t = 9 s      

        Displacement, s = 0

     Substituting

                      s = ut + 0.5 at²

                      0 = u x 9 + 0.5 x -9.81 x 9²

                      u = 44.15 m /s

      Initial velocity is 44.15 m/s

Now we need to find speed when displacement is -23 m

We have equation of motion v² = u² + 2as

     Initial velocity, u = 44.15 m/s  

     Acceleration, a = -9.81 m/s²  

     Final velocity, v = ?

    Displacement, s = -23 m

Substituting  

     v² = u² + 2as

    v² = 44.15² + 2 x -9.81 x -23

    v = 49 m/s  

Distance traveled before stopping is 54.45 m

Speed of the rock just before it reaches the water 23.0 m below the point where the rock left your hand is 49 m/s

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