In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 16.0 m/s

Question

In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 16.0 m/s at an angle of 45.0 below the horizontal. The batter hits the ball toward center field, giving it a velocity of 32.0 m/s at 30.0 above the horizontal. (Take the +x axis to be toward the pitcher and the +y axis to be upward.) (a) Determine the impulse applied to the ball. Ans:___Ns (x component) Ans:___Ns (y component) (b) If the force on the ball increases linearly for 4.00 ms, holds constant for 20.0 ms, and then decreases to zero linearly in another 4.00 ms, what is the maximum force on the ball. Ans:___N (x component) Ans:___N (y component)

Explanation / Answer

I=change in momentum=.2*(32cos30+16cos45)i+.2(32sin30-16cos45)j I=7.85i+2.05j Nsec Ix=7.85Nsec Iy=2.05Nsec b) F=I*4ms=31.4i+8.2j mN Fx=31.4mN Fy=8.2mN

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