In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 11.00 m/
ID: 1789982 • Letter: I
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 42.0 m/s at 30.0° above the horizontal.
(a) Determine the impulse delivered to the ball: i hat + j (Assume that the x-axis is positive toward center field and the y-axis is positive upward.)
(b) If the force on the ball increases linearly for 4.00 ms, holds constant for 20.0 ms, then decreases to zero linearly in another 4.00 ms, what is the maximum force on the ball?
Explanation / Answer
Here ,
initial velocity , u = 11 * (- cos(45) i - sin(45) j)
u = -7.8 i - 7.8 j m/s
final velocity = 42 * (cos(30) i + j * sin(30))
v = 36.4 i + 21 j m/s
a) impulse delivered = m * (v - u)
impulse delivered = 0.20 * ((36.4 i + 21 j) -(-7.8 i - 7.8 j) )
impulse delivered = 8.84 i + 5.76 j Kg.m/s
b)
let the maximum force is F
F * 0.004 * 0.50 + F * 0.004 * 0.50 + 0.012 * F = sqrt(8.84^2 + 5.76^2)
solving for F
F = 659 N
the maximum force is 659 N