The acceleration of a baseball pitcher’s hand as he delivers a pitch is extreme. For a professional player, this acceleration phase lasts only 50 ms, during which the ball’s speed increases from 0 to about 90 mph, or 40 m/s.
What is the force of the pitcher’s hand on the 0.145 kg ball during this acceleration phase? Express your answer with the appropriate units.
Answer:
116 N.
Explanation:
Given,
mass of the ball, m = 0.145 Kg
initial speed = 0 m/s
Final speed = 40 m/s
time = 50 ms = 0.05 s
Force is equal to change in momentum per unit time
[tex]F = \dfrac{m\Delta v}{\Delta t}[/tex]
[tex]F = \dfrac{0.145(40-0)}{0.05}[/tex]
F = 116 N
Force of the pitcher hand on the ball is equal to 116 N.