Afriend throws a baseball horizontally. he releases it at a height of 2.0 m and it lands 21 m from his front foot, which is directly below the point at which he released the baseball. (a) how long was it in the air? (b) how fast did he throw it?

Explanation:

a)Consider the vertical motion of baseball

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = ?

         Displacement, s = 2 m      

     Substituting

                      s = ut + 0.5 at²

                      2 = 0 x t + 0.5 x 9.81 x t²

                      t = 0.64 seconds

The base baseball was in air for 0.64 seconds.

b)Consider the horizontal motion of baseball

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = ?

        Acceleration, a = 0 m/s²  

        Time, t = 0.64 s      

         Displacement, s= 21 m

     Substituting

                      s = ut + 0.5 at²

                      21 = u x 0.64 + 0.5 x 0 x 0.64²

                      u = 32.89 m/s

      Speed of throwing is 32.89 m/s

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