Physics: Impulse and Force?

Question

In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 11.00 m/s at an angle of 45.0° below the horizontal. The batter hits the ball toward center field, giving it a velocity of 46.0 m/s at 30.0° above the horizontal.

(a) Determine the impulse delivered to the ball.
_____ N·s (magnitude)
_____ ° (above the horizontal)

(b) If the force on the ball increases linearly for 4.00 ms, holds constant for 20.0 ms, then decreases linearly to zero in another 4.00 ms, what is the maximum force on the ball?
_____ N (magnitude)
_____ ° (above the horizontal)

Answer 1

a) Impulse is ∆p, where p = momentum (m*v). Find the x and y components of ∆p (∆px = m*∆vx, ∆py = m*∆vy). Vector sum them to get impulse magnitude and direction

b) Impulse is also ∫F*dt. Compute the area under the curve of F vs t (a trapezoid with peak value Fp -- use geometry to compute the area in terms of Fp). Set that equal to the impulse and solve for Fp. The force has the same direction as the impulse.

EDIT: If you need more detail (but no numeric answers!) look at this:

http://img125.imageshack.us/img125/9827/softballimpulseoc7.png