Impulse problem (intro mechanics)

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The discussion revolves around calculating the impulse experienced by a rubber ball after it bounces off the floor and the duration of contact with the floor. The initial speed of the ball before impact is determined to be 5.8 m/s using the formula for free fall. The impulse, calculated as the change in momentum, is found to be 2.4 kg·m/s. Additionally, the time of contact with the floor is computed to be 0.14 seconds based on the average force applied. The poster is seeking clarification on the accuracy of these calculations due to discrepancies reported by their electronic assignment system.
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Homework Statement



A rubber ball (mass 0.26 kg) is dropped from a height of 1.7 m onto the floor. Just after bouncing from the floor, the ball has a speed of 3.6 m/s.
(a) What is the magnitude and direction of the impulse imparted by the floor to the ball? Magnitude ? kg · m/s

(b) If the average force of the floor on the ball is 17 N, how long is the ball in contact with the floor?


Homework Equations



dp=Fdt=m(vf-vi)

The Attempt at a Solution



a)the unknown is the initial speed of the dropping ball under free fall:

vi= sqrt (2 x 9.8 x 1.7) = 5.8

delta p= 0.26 (3.6--5.8)=2.4

b) dt=dp/Fave=0.14 s

Please let me know what is wrong here, my electronic assignment system is telling me the numbers are wrong. Thank you!
 
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Your answers all seem right to me.
 
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