A baseball 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.