Question

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, 8.00 s after it was thrown. What is the speed of the rock just before it reaches the water 25.0 m below the point where the rock left your hand?

Answer 1

The speed of the rock just before it reaches the water 25.0 m below the point where the rock left your hand  is 45.06 m/s

Explanation:

First let us find the initial velocity,

We have after 8 seconds the displacement is zero,

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = ?

        Acceleration, a = -9.81 m/s²

        Time, t = 8 s      

        Displacement,s = 0 m

     Substituting

                      s = ut + 0.5 at²

                      0 = u x 8 + 0.5 x -9.81 x 8²

                      u = 39.24 m/s

Initial velocity is 39.24 m/s.

Now this case is similar to case where a rock is thrown at 39.24 m/s downward.

We have equation of motion v² = u² + 2as

      Initial velocity, u = 39.24 m/s  

      Acceleration, a = 9.81 m/s²  

      Final velocity, v = ?

      Displacement, s = 25 m  

     Substituting  

        v² = u² + 2as

        v² = 39.24² + 2 x 9.81 x 25

        v = 45.06 m/s

The speed of the rock just before it reaches the water 25.0 m below the point where the rock left your hand  is 45.06 m/s