In a slow-pitch softball game, a 0.200-kg softball crosses the plate at 13.0 m/s
ID: 1355021 • Letter: I
Question
In a slow-pitch softball game, a 0.200-kg softball crosses the plate at 13.0 m/s at an angle of 40.0° below the horizontal. The batter hits the ball toward center field, giving it a velocity of 50.0 m/s at 30.0° above the horizontal. (Take positive i to be the horizontal direction from the plate toward center field and take positive j to be the upward vertical direction.)
(a) Determine the impulse delivered to the ball. I with arrow = 10.64i+6.67j Correct: Your answer is correct. N · s
(b) If the force on the ball increases linearly for 4.00 ms, holds constant for 20.0 ms, and then decreases linearly to zero in another 4.00 ms, what is the maximum force on the ball?
Explanation / Answer
impulse = change in momentum
in horizontal direction ,change in momentum=impulse = (13cos40 + 50 cos 30 ) * 0.200 =10.65196959
in vertical (j) = 13*sin 40+50*sin 30 * 0.200 = 6.67124778519
(b) area under Force time curve gives total impulse so
1/2 *Fmax *4 + 20*Fmax +1/2 *Fmax *4 =
24Fmax * 10^-3 = 12.5578063371
Fmax =523.241930712 Newton