During a throw, a baseball pitcher exerts a force of 3.20 m before releasing it. If the 0.12-kg ball leaves the pitcher's hand with a speed of 35.0 m/s, how much of a force did the pitcher exert on the ball?
2006-10-03 12:59:23 · 2 answers · asked by Bao Wow 3 in Science & Mathematics ➔ Physics
During a throw, a baseball pitcher exerts a force on the ball though a distance of 3.2 m before releasing it. If the 0.12-kg ball leaves the pitcher's hand with a speed of 35.0 m/s, how much of a force did the pitcher exert on the ball?
sorry the question was mistyped
2006-10-03 13:28:23 · update #1
In 3.2m the ball reaches 35.0 m/s velocity from zero. Additionally, we know that the pitcher axerts force to the ball. Therefore, there have to be acceleration
First, we calculate the acceleration:
V^2-V0^2=2*a*x
(35.0)^2-0=2*a*3.2
a=191.41 m/s^2
To calculate force:
F=ma
F=0.12*191.41
F=22.97N
2006-10-03 18:49:28 · answer #1 · answered by Farshad 2 · 0⤊ 0⤋
Its been awhile since physics class
but if F = ma
then .12kg * 35m/s = 4.2
2006-10-03 20:03:09 · answer #2 · answered by jbscooby99999 3 · 0⤊ 1⤋