In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 18.0 m/s
ID: 1458909 • Letter: I
Question
In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 18.0 m/s at an angle of 40.0° below the horizontal. The batter hits the ball toward center field, giving it a velocity of 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)
Explanation / Answer
change in momentum along forward direction
dPx = m v cos 30 - mu cos 40
dPx = 0.2 * 46 * cos 30 - (0.2 *18 * cos 40)
dPx = 5.21 kg m/s
Change in momentum along y axis
dPy = m v sin 30 -(- mu sin 40 )
dPy = 0.2 * 46 *sin 30 + (0.2 *18 * sin 40)
dPy = 6.91 kgm/s
Net Momentum change = Impulse = F t
Net Momentum change = P^2 = Px^2 + Py^2
P^2 = 5.21^1 + 6.91^2
P = 7.27 Kgm/s
direction tan theta = Py/Px
tan theta = 6.91/5.21
theta = 53 deg
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max force Fmax = J/t = 7.27/4 e-3
Fmax = 1817.5 N
angle 53 deg