Question

6. A baseball player hits a 0.155-kilogram fastball traveling at -44.0 m/s into center field at a speed of 50.0

m/s. If the impact lasts for 0.00450 second, with what force does he hit the

baseball?

Answer 1

3,237.78N

Step-by-step explanation:

According to Newton's second law of motion

F = mass * acceleration

Since a = v-u/t

F = m(v-u)/t

Given

Mass m = 0.155kg

v = 50.0m/s

u = -44.0m/s

Time t = 0.00450secs

Substitute

F = 0.155(50-(-44))/0.00450

F = 0.155(50+44)/0.00450

F = 0.155(94)/0.00450

F = 14.57/0.00450

F = 3,237.78N

hence he hit the baseball with a force of 3,237.78N