Abatter hits a foul ball. the 0.140-kg baseball that was approaching him at 40.0 m/s leaves the bat at 30.0 m/s in a direction perpendicular to the line between the batter and the pitcher. what is the magnitude of the impulse delivered to the baseball?

Given in the question,

mass of foul ball = 0.140 kg

initial speed with which ball was hit with the bat = 30 m/s

final speed  = 40 m/s

According to the scenario the whole scene is making a right angle triangle

So, to the solve the question we will use pythagorus theorem

Here,

Hypotenuse= Magnitude of impulse

Base = 1st change of momentum

height = 2nd change of momentum

1st impulse (1st change of momentum)

p = m(1)v(1) = (0.14 kg)(40.0 m/s) = 5.6 kg m / s = 5.6 N s

2nd impulse (2nd change of momentum)

p = m(2)v(2) = (0.14 kg)(30.0 m/s) = 4.2 kg m / s = 4.2 N s

Magnitude of impulse (hypotenuse of triangle)

impulse² = (5.6)² + (4.2)²

impulse² = 31.36 + 17.64

impulse² = 49

impulse² = √49

impulse = 7.0 N s

d.7.5cm is your answer