Abaseball player swings his bat at a pitch and manages to hit it with a velocity of 60 mph as it leaves the bat. the ball has an initial angle of 55 degrees to the horizontal and he hits the ball at a vertical location 80 cm off the ground. how far horizontally (the "range") does the ball travel before before hits the ground ?

The ball travels 27.15 m horizontally before hitting the ground.

Explanation:

Considering the vertical motion of baseball ( upward positive), we need to find when it reaches the ground

      Initial velocity, u = 60 sin 55 = 49.15mph = 13.65 m/s

      Acceleration, a = -9.81 m/s²

      Displacement, s = - 80 cm = -0.8 m

Substituting in s = ut + 0.5 at²

     -0.8 = 13.65 x t - 0.5 x 9.81 x t²

        4.905 t² -13.65 t - 0.8 = 0

       t = 2.84 s or t = -.057 s ( not possible )

So the ball will hit ground after 2.84 s

We need to find horizontal distance traveled during 2.84 seconds

       Initial velocity, u = 60 cos 55 = 34.41mph = 9.56 m/s

      Acceleration, a = 0 m/s²

      Time, t = 2.84

Substituting in s = ut + 0.5 at²

     s = 9.56 x 2.84 + 0.5 x 0 x 2.84²

     s = 27.15 m

The ball travels 27.15 m horizontally before hitting the ground.        

show the diagram then i will answer