A 0.145-kg baseball pitched horizontally at 31.5 m/s strikes a bat and is popped straight up to a height of 59.0 m .

Question

A 0.145-kgkg baseball pitched horizontally at 31.5 m/sm/s strikes a bat and is popped straight up to a height of 59.0 mm .

If the contact time is 1.80 msms , calculate the magnitude of the average force on the ball during the contact.

Express your answer using two significant figures

Answer 1

Let

m = 0.145 kg be the mass of the baseball,

v = 31.5 m/s be the horizontal speed of the baseball,

w (unknown) be the vertical speed of the baseball as it leaves the bat,

h = 59 m be the height the baseball reaches,

t = 1.8 ms = 0.0018 s be the contact time,

g = 9.81 m/s² be gravitational acceleration,

F be the force to be calculated.

The force F has a horizontal component F_h causing the ball to decelerate

horizontally from v to 0.

And the force F has a vertical component F_v causing the ball to accelerate

vertically from 0 to w.

The horizontal deceleration is v/t, so by Newton's second law

F_h = mv/t

The vertical acceleration is w/t, so by Newton's second law

F_v = mw/t

It remains to calculate w, which we can do from conservation of energy:

The kinetic energy after striking the ball gets converted to the

potential energy at the height h:

mw²/2 = mgh

w = √(2gh)

The total force F is the vector sum of F_h and F_v, so its magnitude is

|F| = √(|F_h|² + |F_v|²) = √(m²v²/t² + 2m²gh/t²) = (m/t)×√(v² + 2gh)

Substituting actual numbers

|F| = (0.145/0.0018)×√(31.5² + 2×9.81×59) = 3735 N