-r16 points In a slow-pitch softball game, a 0.200 kg softball crossed the plate
ID: 2032646 • Letter: #
Question
-r16 points In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 10.00 m/s at an angle of 35.0° below the horizontal. The batter hits the ball toward center field, giving it a velocity of 40.0 m/s at 30.0° above the horizontal My Notes (a) Determine the impulse delivered to the ball: 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? (magnitude) ° (above the horizontal)Explanation / Answer
Given data
mass m =0.2 kg, velocty u =10 m/s at 35 deg
velocity v =40 m/s at 30 deg
x-components for change in momentum,
?Px =mvcos30-mucos35 =5.289 kg.m/s
y-components for change in momentum,
?Py =mvsin30-(-musin35) =mvsin30+(musin35)=5.14 kg.m/s
Impule I= 5.289i + 5.14j
B.
magnitude of the change in momentum is
?P =sq.rt[(?Px)^2+(?Py)^2] =7.38 kg.m/s
Calculate the force
F = impluse/time= 7.38/20*10^-3 = 369 N
Angle= tan^-1(5.14/5.289)= 44.81
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