A(n) 0.155 kg baseball is thrown with a speed of 18.4 m/s. it is hit straight back at the pitcher with a final speed of 18.4 m/s. what is the magnitude of the impulse delivered to the ball? answer in units of kg · m/s.

5.704kg.m/s toward the pitcher

Explanation:

Impulse (Ft)= change in momentum

Impulse= m(v-u)

m=0.155 kg  ,       v=18.4 m/s  , u=18.4m/s

impulse=0.155[-(18.4) - (18.4)]

        =  0.155(-36.8)

        = -5.704 kg.m/s

the negative means towards the pitcher

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

not too sure thinking all accelerations are pointed down