Need help with B In a slow-pitch softball game, a 0.200 kg softball crossed the

Question

Need help with B

In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 16.00 m/s at an angle of 35.09 below the horizontal. The batter hits the ball toward center field, giving it a velocity of 42.0 mis at 30.0° above the horizontal. (a) Determine the i 9.8956 Ns603544 Ns(Assue that the x axis is positive toward center field and the y axis is posidve upward.) (b) If the force on the ball increases linearly far 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? Enter a number. (magnitude) 31.37(above the horizontal)

Explanation / Answer

This can be calculated as area under the graph for Force vs time graph.

I = F*T

F = I/T

where T is toal time , T = 4+20+4 = 28 ms = 0.028 sec

so,

F = 9.8956 i + 6.03544 / 0.028

F = 353.41 i + 215.55 j

Magntitude of F = 413.95 N

direction = arctan (215.55 / 353.41)

direction = 31.379 degrees

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